if you have a regression is it in general true that the inaccuracy trying to be minimized will, if a new data point is added, be minimized the most if the new point exactly follows the regression on the rest of the points
#general discussion
sour sinew
dusty flax
ohai, can i ask a math question here? not related to school or university. Just need a little knowledge
agile roost
ohai, can i ask a math question here? not related to school or university. Just ...
go ahead
dusty flax
I want to "translate" this expression into mathematical (graduated long time ago and forgot everything):
It's like sum sequence / limit which contains itself. At infinity it should be 0.2, but idk how to write it right
sour sinew
I want to "translate" this expression into mathematical (graduated long time ago...
isnt that just 1/6 + 1/36 + 1/216 + 1/1296...
finite sum of geometric series
multiplying it by 6 we gain a +1 and lose a +1/1296 meaning 6x = x + 1 - 1/6^4 meaning x = (1-1/6^4)/5
and you can see that indeed limits to 1/5
dusty flax
isnt that just 1/6 + 1/36 + 1/216 + 1/1296...
i'm ashamed lol. Thought its smth more difficult
lean marsh
Hi there people, looking for someone who's good at combining and calculating stuff like this:
quantity: 18.000 - Boar man, 85% fire damage | 95% ice damage
quantity: 16.000 - Liodile, 110% fire damage | 105% ice damage
quantity: 13.000 - Carnivostrich, 110% fire damage | 110% ice damage
In this game, I can deal either ICE or FIRE damage.
I'm wondering which damage to use for maximum damage over time. For example the Boar Man takes 15% reduced fire damage, and 5% reduced ice damage.
Boar Man, Liodile and Carnivostrich all take damage at the same time by AOE dmg, (area of effect).
Have fun, I will follow this thread to see what people come up with. I'm too bad to calculate the actual values with quantity vs % dmg etc 😄
radiant urchin
Ik it’s a little late but this book is incredible, totally recommend for those who havnt read it!
turbid vine
Was reading this paper (http://dfns.dyalog.com/n_ratrep.htm) cause it was linked in a stack exchange article, and it said Euler claimed:
Where:
S(x) = ... + x∧(3) + x∧(2) + x∧(1) + x∧(0) + x∧(-1) + x∧(-2) + x∧(-3) + ...
Then:
S(x) = 0
I can't find anything about this online, does anyone have a link to any resource about this? Or is this not real?
dire sable
Ik it’s a little late but this book is incredible, totally recommend for those w...
Nice to hear you enjoyed it too! It's a really good book that introduces a lot on interesting concepts and ideas
velvet cosmos
I've been really interested in all of the number classifications and want to make a program using them. Does anyone know a list of them all? I know there are probably thousands of them but if anyone has a general list of some of them it would be really helpful. And by number classifications, I mean motzkin numbers, bell numbers, tribonacci numbers. Like all of those obscure number classifications.
sour sinew
I've been really interested in all of the number classifications and want to mak...
oeis
versed comet
can someone help
ive tried u sub. I briefly considered completing the square before realising the 9/2y would be interfering with that
I tried thinking of a possible trig sub
none that I can think of
quick drum
hiiiiii
quick drum
its easy
vivid gulchBOT
ancient helm
Hmmm
dusty swallow
hmmm
vivid gulchBOT
sour sinew
Da f
it just converted it into a rational function,
factored the bottom, and did partial fractions
and integrated each
ancient helm
Most cool
agile roost
and integrated each
such a goofy integral
dusty swallow
0
agile roost
nope!
bro didn't know Poland is FULL of poles
idiot!!
@hollow ice you seeing this??
agile roost
No, you're an idiot either way
jokes are meant to be funny, you're the only exception
dusty swallow
bros mad that he missed it
agile roost
you literally failed miserably
idiot!
not a funny joke anyway Mr German
I'm not allowed to tell you to follow in the late Fuhrer's footsteps
dusty swallow
Bro really is going on a rampage
hollow ice
what function are we integrating bruh
trim valley
pll is the last step of solving a rubik cube, where last layer must be solved and all peices have same orientation
it has 21 cases
but if i do math i get 4!^2/2/4=72
(4!^2 for 4! for the corners and 4! for the edges divide 2 for odd number of edge or corner swap to exlude impossible and finally divide 4 so color doesnt matter)
i know it is because some are mirrors of each other or they are different ways of being the same
how can this be expressed as formula to get 21 instead (using logic)
agile roost
what function are we integrating bruh
integrating POLAND over IT'S BORDER
hollow ice
integrating POLAND over IT'S BORDER
Oh.
ancient helm
Wouldn’t that be the land area of Poland
Or total area ig
Now find the volume when revolving it around the equator
dusty swallow
Wouldn’t that be the land area of Poland
💀
ancient helm
Is it bad that idk what a contour integral is
sour sinew
wow 1 hour exactly
*3607.236 seconds
ancient helm
Still really close to 1 hour
solar marlin
Hi maths ppl
ancient helm
Hey
marsh imp
Anyone know how I can show that n!/(n^n) is decreasing aka a_(n+1) < a_n ?
ancient helm
Inductive proof perhaps
marsh imp
Is there any easier method?
solar marlin
Inductive proof would be easiest ngl
solar marlin
do u need to show that its decreasing as n increases?
marsh imp
2/n^2 going to zero automatically implies that as n gets larger 2/n^2 gets smaller (decreases), and since n!/n^n is smaller, it also goes to zero and n goes to infinity, therefore a_n+1 < a_n??
ancient helm
So (n+1)!/((n+1)^(n+1)) < (n+1)! / ((n+1) • n^n) = n! / (n^n
That’s what I would do
Basically show that (n+1)!/((n+1)^(n+1)) < n!/n^n
In latex form it would be:
$\frac{(n+1)!}{(n+1)^{n+1}} < \frac{(n+1)!}{(n+1)•n^n = \frac{n!}{n^n}$
vivid gulchBOT
r•eⁱˣ = r•cos(x) + ri•sin(x)
$\frac{(n+1)!}{(n+1)^{n+1}} < \frac{(n+1)!}{(n+1)•n^n = \frac{n!}{n^n}$
```Compilation error:```! File ended while scanning use of \frac .
<inserted text>
\par
<*> 858341562143735860.tex
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```
solar marlin
L
dusty swallow
L
ancient helm
Dam
dusty swallow
solar marlin
Can I show u lot my maths pls
Probably not impressive bc im only 17 and likely really bad compared to most of u here
ancient helm
Go ahead
solar marlin
Some implicit differentiation i did today
ancient helm
Nice
dusty swallow
very cool
marsh imp
hmm how will this help me?
solar marlin
Doing an easy trig integral by substitution for the challenge
ancient helm
I have a_n = n!/n
Yea and dappy showed that a(n+1) < a(n)
Which is the same as it decrasing
marsh imp
Do u mind explaining how its being showed that a(n+1) < a(n)
ancient helm
Read dappy’s proof
dusty swallow
its my cursor
marsh imp
oh lol
solar marlin
Do u mind explaining how its being showed that a(n+1) < a(n)
surely just say:
if n∈z, n⊆(n+1) => n+1>n ∴a(n+1)>a(n), ∀n∈z
sorry if my notation is bad
Again I'm not some maths god
marsh imp
it literally shows what u asked for
How did u get from the first line to the second?
dusty swallow
but am not sure what's being done...
all im really doing is canceling the n+1 and then since n+1 is greater than n, (n+1)^n is greater than n^n, and when the denominator is bigger its smaller
solar marlin
surely just say:
if n∈z, n⊆(n+1) => n+1>n ∴a(n+1)>a(n), ∀n∈z
Surely this holds up
Oh nvm I read wrong
marsh imp
all im really doing is canceling the n+1 and then since n+1 is greater than n, (...
1 (2) ..... (n+1) is the same as (n+1)! right?
ancient helm
1 (2) ..... (n+1) is the same as (n+1)! right?
Yes that’s the definition of factorial
sour sinew
and it's also true from definition of sequence
ancient helm
n! = n•(n-1)•(n-2)…•3•2•1
ancient helm
and it's also true from definition of sequence
Which is true from definition of definition
sour sinew
definition of true
marsh imp
So then dappy has shown that if a_n = n!/(n^n), then a_ (n+1) < a_n for all natrual numbers or only n >= 1?
sour sinew
Basically show that (n+1)!/((n+1)^(n+1)) < n!/n^n
yeah definitely just divide both sides by n! then simplification happens
ancient helm
definition of true
true = !false
sour sinew
no induction necessary
ancient helm
So then dappy has shown that if a_n = n!/(n^n), then a_ (n+1) < a_n for all natr...
For all natural numbers n
As it works regardless of what we put for n
Now prove for real numbers
sour sinew
So then dappy has shown that if a_n = n!/(n^n), then a_ (n+1) < a_n for all natr...
0^0 is not defined
so
marsh imp
no induction necessary
Yeah I was trying to show by induction rn but it's so pain
sour sinew
only n >= 1
marsh imp
0^0 is not defined
that's 1?
sour sinew
normally assigned the value 1
marsh imp
In algebra are we allowed to just take the facotrial of both sides of an equation?
sour sinew
yeah but why would you do that
marsh imp
well i'm tryna show n!/(n^n) decreases by induction
sour sinew
yes
solar marlin
Yes that’s the definition of factorial
I found out that n!=n(n-1)! In my head when I was like 12 and thought I was smart
ancient helm
I found out that n!=n(n-1)! In my head when I was like 12 and thought I was smar...
Same
sour sinew
amazing
marsh imp
u guys r legends
ancient helm
I also found out the (x+y)^2 formula and thought I was smart
ancient helm
Divide by r :)
ancient helm
Yes
solar marlin
I realised
ancient helm
Who doesn’t love Euler
sour sinew
realized cos(pi) = -1 no way
ancient helm
realized cos(pi) = -1 no way
Ain’t no way bruv
sour sinew
impossible
ancient helm
I thought it was 12
solar marlin
Sqrt(e^iπ)=i
So rewrite e^(iπ) but every time i appears rewrite it as sqrt(e^iπ)
How do you write that
sour sinew
cos(pi) = -1/12
solar marlin
How do you write that
It just goes on forever
solar marlin
yea that
marsh imp
ty for the help guys!
sour sinew
So e^(e^(e^…ipi/2)ipi/2… ?
nah should be e^(π/2e^(π/2e^(π/2...))))
clearly remains i as you go off to infinity
real
solar marlin
Do you know what fractals are?
No yes
|
|
|
Do you now? > yes no
|
|
|
|
Do you now? .....
there's a meme for that but I lost it :(
sour sinew
im god
sour sinew
search the internet, folks
solar marlin
Lol
marsh imp
n!/(n^n) > 0 for n >=1 right?
marsh imp
ye ty
dusty swallow
This is quite simple
Let’s say we compute the first 5 terms of the infinite series
Then the error to the real value of the infinite series is determined by all of the series members n>5
Notice that our series is a part of n! / n^n multiplied by sin something
To achieve the greatest error, sin would have to be constantly minus 1 or 1
More formally we can create the upper bound n!/n^n for the absolut error
But we know series n! / n^n = e (no we don’t know this tf am I smoking, n is not x u moron)
What
Omfg
I read the whole problem wrong lmao
sour sinew
Notice that our series is a part of n! / n^n multiplied by sin something
where did you get sin
n!/n^n = e
marsh imp
Anyone know how I can solve for k here?
sour sinew
is that for sum of an alternating seried
marsh imp
ye
sour sinew
also, the way to solve for it is with a calculator
though you could also just see this is true for k = 100 or something
lethal wigeon
hmm
open whale
@ashen agate @dusty swallow
Let there be a polynomial $p_c(z)$ of degree greater than $1$.\
Then, for the sequence of polynomials defined by $z_0(c)=0$, and $z_{n+1}(c)=p_c(z_n(c))$, there exists a polynomial $h_{n,k}(c)$ such that $z_{nk}(c)=h_{n,k}(c)\cdot p_c(z_n(c))+z_n(c)$
vivid gulchBOT
yoavmal
open whale
I think this is true, I am not 100% confident in this
The proof is using two inductions one after another
Same I used for the specific case with z²+c
I'll be trying to do that in the soon time
Formally
open whale
$p_c(z)=\sum_{n=0}^da_n(c)z^n$ where $a_n(c)$ is a polynomial of $c$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
so $a_0(c)$ should be a polynomial of degree $1$ in terms of $c$, at least
vivid gulchBOT
yoavmal
open whale
next up, $z_2=p_c(a_0(c))$
vivid gulchBOT
yoavmal
dusty swallow
<@635925298243829782> <@468053685805514752>
@dusty swallow
open whale
<@468053685805514752>
do you agree
vivid gulchBOT
dappy
dusty swallow
This means $z_n(x)=x^{(2^n)}$
vivid gulchBOT
dappy
dusty swallow
Ur statement would say there exists a polynomial $h_{n,k}$ such that
vivid gulchBOT
dappy
dusty swallow
$x^{2^{kn}} = h_{n,k} \cdot x^{2^{n+1}} + x^{2^n}$
vivid gulchBOT
dappy
dusty swallow
@open whale
open whale
Ur statement would say there exists a polynomial $h_{n,k}$ such that
I added further specifications down the line
dusty swallow
Like
open whale
a₀ must be nonezero
To be precise, it must be a polynomial of c with degree 1 or more
open whale
Back to the proof
So we have that
$z_2(c)=p_c(z_1(c))=\sum_{n=0}^da_n(c)(z_1(c))^n=a_0(c)+a_1(c)z_1(c)+\sum_{n=2}^da_n(c)(z_1(c))^n$
vivid gulchBOT
yoavmal
open whale
What do I do with that thing lmao
Lets see what did I do with the specific proof
Lets show to with z₁
$z_1(c)=p_c(0)=a_0$
vivid gulchBOT
yoavmal
vague dagger
Yoav lol
dusty swallow
First off I want to write
Bro casually died
ancient helm
And now I will prove:
halcyon heart
Is it mathematical incorrect to have two equal signs on the same line? Some of my teachers say it is, but I can't find any source anywhere except for a few quora answers
dusty swallow
Do u mean something like 3*2+3=6+3=9?
halcyon heart
Yes
dusty swallow
Well if u have something like x^2 =3x u can’t directly say =x=3 that is obv false
But the equation from before “implies” the second one
That’s what the arrows mean
dusty swallow
like, r=20/4=>5
Nah there u can just aswell use an equal sign
Actually the way it’s written right there is prob on a formal level even wrong
Better to use an equal sign in that case
open whale
Bro casually died
i had to go lmao
uhh
so we have the base case
now we want the induction
lets assume
$z_{n+k}(c)=h\cdot p_c(z_n(c))+z_k(c)$
vivid gulchBOT
yoavmal
open whale
then
you know what
i'm removing the Cs
it's going to make it a little too messy now and so more prone to errors
$z_{n+k}=h\cdot p(z_n)+z_k$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
so uhh
$=\sum_{m=0}^da_m(h\cdot p(z_n)+z_k)^m$
now, what i want to do, is somehow extract z(k+1) from this
which is p(z(k))
wait this is wrong
this is better
so we can break it down to the binomial
but i don't want to bother
uhh
actually i will, why not
oh also another issue
i used n as an index too
vivid gulchBOT
yoavmal
open whale
now we can do binomial
i suppose over j now?
$=\sum_{m=0}^da_m\sum_{j=0}^m\binom{m}{j}(hp(z_n))^jz_k^{m-j}$
vivid gulchBOT
yoavmal
open whale
now we want to extract the z(k+1) polynomial
so if j=0 we get z^m
that's not what we want
oh
it is
because m moves between 0 and d
so that'd be our polynomial
$=\sum_{m=0}^d\qty(a_mz_k^m+a_m\sum_{j=1}^m\binom{m}{j}(hp(z_n))^jz_k^{m-j})$
vivid gulchBOT
yoavmal
open whale
$=\sum_{m=0}^da_mz_k^m+\sum_{m=0}^da_m\sum_{j=1}^m\binom{m}{j}(hp(z_n))^jz_k^{m-j}$
vivid gulchBOT
yoavmal
open whale
$=p(z_k)+\sum_{m=0}^da_m\sum_{j=1}^m\binom{m}{j}(hp(z_n))^jz_k^{m-j}$
vivid gulchBOT
yoavmal
open whale
$=p(z_k)+\sum_{m=0}^d\qty(a_m(hp(z_n))^m+a_m\sum_{j=1}^{m-1}\binom{m}{j}(hp(z_n))^jz_k^{m-j})$
vivid gulchBOT
yoavmal
open whale
$=p(z_k)+\sum_{m=0}^da_m(hp(z_n))^m+\sum_{m=0}^da_m\sum_{j=1}^{m-1}\binom{m}{j}(hp(z_n))^jz_k^{m-j}$
vivid gulchBOT
yoavmal
open whale
$=p(z_k)+p(hp(z_n))+\sum_{m=0}^da_m\sum_{j=1}^{m-1}\binom{m}{j}(hp(z_n))^jz_k^{m-j}$
vivid gulchBOT
yoavmal
dusty swallow
Bro is having a stroke and thought I wouldn’t notice
open whale
i don't need that part
$=p(z_k)+z_n\sum_{m=0}^da_m\sum_{j=1}^m\binom{m}{j}(hp(z_n))^{j-1}z_k^{m-j}$
vivid gulchBOT
yoavmal
open whale
Bro is having a stroke and thought I wouldn’t notice
i may have underestimated this
sorry
not underestimated
uhh
what's the word
miscalculated
it is possible i have phrased the original thing wrong
and it's supposed to be
$z_{n+k}=p(hz_n)+z_k$
vivid gulchBOT
yoavmal
open whale
and it's possible i phrased the entire thing wrong
yeah, definitely
$=p(z_k)+\sum_{m=0}^da_m\sum_{j=1}^m\binom{m}{j}(hp(z_n))^jz_k^{m-j}$
vivid gulchBOT
yoavmal
open whale
so this is definitely correct
$=z_{k+1}+\sum_{m=0}^da_m\sum_{j=1}^m\binom{m}{j}(hp(z_n))^jz_k^{m-j}$
vivid gulchBOT
yoavmal
open whale
now the rest needs to be written in some form of zn
i mean, i can definitely factor the zn out of this
but this isn't what i want
oh wait i'm stupid
it's right there
$=z_{k+1}+p(z_n)\sum_{m=0}^da_m\sum_{j=1}^m\binom{m}{j}(hp(z_n))^{j-1}z_k^{m-j}$
here we go
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
and $z_{0+1}=0p(z_0)+z_1$
vivid gulchBOT
yoavmal
open whale
and alternatively
actually this is not general enough
the proof for induction assumes more
here we go
proof
$z_0=0$
vivid gulchBOT
yoavmal
open whale
$z_{n+0}=1\cdot z_n+z_0$
vivid gulchBOT
yoavmal
open whale
this is the base for the induction
is it valid?
@dusty swallow
yeah i don't think i need a second check for that, it's correct
and for extra measure we also can prove
you know what
nvm
i'll work on it a little more
i want to prove first that for every n,k
there exists a polynomial h so that
$z_{n+k}=h\cdot p(z_n)+z_k$
vivid gulchBOT
yoavmal
open whale
now, what seems odd is this doesn't hold precisely for k=0
nor for k=1
uhh
$z_{n+0}=0p(z_0)+z_n$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
this willl take some while
sour sinew
the crackpot team moves letters around until they find the right combination
sour sinew
Let there be a polynomial $p_c(z)$ of degree greater than $1$.\\
Then, for the s...
i assume you meant z0(c) = c cuz otherwise it's constants in the whole sequence, and that this works for all posints n, k > 1?
open whale
i assume you meant z0(c) = c cuz otherwise it's constants in the whole sequence,...
z0=0
z1=c if p(z)=z²+c
We assume a0 in our polynomial is of degree 1 in terms of c
sour sinew
z1=c if p(z)=z²+c
that is still constant
maybe im misunderstanding
i guess it's p(c,z) then the sequence is 0, p(0,c), p(p(0,c),c)...
and h is supposed to work for all values of c
c is normally a constant so 😅
confused
open whale
c is normally a constant so 😅
We treat c as the variable
Well, we play on that iterativeness
The polynomial we talk on is a polynomial of z
With coefficients which are polynomials of c
Such that the degree of a0 is at least 1 in terms of c
burnt nest
,w solve 4x² - 1 = 0
median gate
heyyy guys
ancient helm
What makes math fun for you guys
For me it’s how it feels objective, black/white and logical, unlike something like English or politics which is very gray and has a lot up for debate
agile roost
For me it’s how it feels objective, black/white and logical, unlike something li...
racist much
ancient helm
racist much
The people that genuinely think math is racist for some reason:
agile roost
The people that genuinely think math is racist for some reason:
that's why it's fun
solar quarry
Math is objective
Meanwhile Axiom of Choice, Continuum hypothesis:
agile roost
> Math is objective
Meanwhile Axiom of Choice, Continuum hypothesis:
feels objective
mfs before nothing makes sense
solar quarry
Well, close enough anyway
agile roost
math is cool because this statement is fallacious
dusty swallow
Proof by contradiction
Alex sucks ass
No contradiction found
QED
Most valid mathematical proof Ong
agile roost
Alex sucks ass
that is a contradiction by definition
solar quarry
Math is cool cause it’s creative
Now this is something I could get by (/j ?)
agile roost
Now this is something I could get by (/j ?)
watching walking dead
placid bough
The people that genuinely think math is racist for some reason:
You remind me of an article about racism in math that was actually well written and thought provoking
ancient helm
You remind me of an article about racism in math that was actually well written ...
Send link to article plz
dusty swallow
The point they are making is valid but it’s not racism in math, it’s racism in our culture regarding math
placid bough
Yeah
idle mist
For me it’s how it feels objective, black/white and logical, unlike something li...
Yeah true, but for me, it’s the discussion and collaboration that makes math fun for me
Is it weird😂
Like would you rather do math yourself or discuss and collaborate with others
ancient helm
Hmm I haven’t tried collaborative math before
sour sinew
> Math is objective
Meanwhile Axiom of Choice, Continuum hypothesis:
, whether 0 is natural:
(joke!!)
solar quarry
, whether 0 is natural:
:O you found much better one
sour sinew
i dont think this was the first time this was ever posed
it's weird that they even had to name it after anyone
humans have been looking for shorter road journeys forever
surely
wait, the shortest cycle?
solar quarry
Shortest cycle might have been relatively uncommon
sour sinew
maybe'
candid cedar
, whether 0 is natural:
imo 0 should be a natural number in set theory and computer science but not a natural number in analysis
sour sinew
imo 0 should be a natural number in set theory and computer science but not a na...
does analysis touch N
i need to look into that
solar quarry
Imho It should be natural number in algebra as well
sour sinew
it appears that peano arithmetic explicitly makes 0 natural
solar quarry
(Maybe was a hot take)
sour sinew
whaaaaaaaa
candid cedar
In set theory, 0 is an ordinal but it is not infinite so it should be a natural number
candid cedar
whaaaaaaaa
This is why there is also a difference between omega and N
candid cedar
Like 0 in omega is the empty set, but in N is the set of all rational numbers (in the version that is not a subset of R) less than 0
solar quarry
whaaaaaaaa
Nothing is strange about this though.
candid cedar
Assuming the real numbers are constructed through the dedekind cuts
solar quarry
It is tricky to extend a set satisfying good properties, you better make new one.
You can replace the corresponding elements (e.g. 1 and 1.0) later. (I dislike this step btw, I prefer them to be individual)
sour sinew
Like 0 in omega is the empty set, but in N is the set of all rational numbers (i...
well, 0 in N being the set of all rational numbers less than 0 is a bit beyond me
oh it's just {}
duh
candid cedar
well, 0 in N being the set of all rational numbers less than 0 is a bit beyond m...
Here when I write N I am meaning as a subset of R
You never actually use that the real numbers were constructed this way. It is just that this construction allows us to create a notion of R,+,*,< that satisfies certain axioms. Then we just use those axioms to prove whatever about R
sour sinew
if this construction of R does not help prove anything about R why was it done
just to see if it could be done? if R was possible
sour sinew
yayyy R exists hooray
candid cedar
So this means that if you are doing things with R and calculus and you create a contradiction, ZF is inconsistent and it is the set theorists problem now
solar quarry
It means that R makes sense, there have been many mathematical constructions which turns out to be not working
E.g. you should not do differentiation/integration naively
sour sinew
So this means that if you are doing things with R and calculus and you create a ...
does that mean all real results are cancelled
open whale
@sour sinew want to help me with my search for a general polynomial recursive formula?
i want to describe the iterative sequence of polynomials in a meaningful manner
as in
$z_{n+k}=z_n+z_k$
vivid gulchBOT
yoavmal
open whale
in concept, of course
something along these lines, with stuff done to zn and zk
right now i have
for a sequence $z_{n+1}=p(z_n)$
vivid gulchBOT
yoavmal
sour sinew
is z the polynomial
open whale
where $p(z)=a\cdot z^m+c$
vivid gulchBOT
yoavmal
open whale
and $z_0=0$
vivid gulchBOT
yoavmal
open whale
what i have is
sour sinew
c, ac^m + c, a(ac^m+c)^m + c
open whale
$z_{n+k}=ha\cdot z_n^m+z_{k+1}$
vivid gulchBOT
yoavmal
open whale
c, ac^m + c, a(ac^m+c)^m + c
zn is the Nth term in the sequence
and it behaves as a polynomial with respect to c
open whale
**yoavmal**
where h is some polynomial depending on the particular iteration
sour sinew
is this a generalization of the mandelbrot recurrence
open whale
is this a generalization of the mandelbrot recurrence
yes
i have proven originally that
for the mandelbrot recurrence
we get
$z_{n+k}=h\cdot z_n^2+z_{k+1}$
vivid gulchBOT
yoavmal
open whale
wait let me re-prove it real quick
we have
wait
uhh
let me remember how it was phrased precisely
sour sinew
so it's really function from ordered pairs to ordered pairs
$a_{n-1} = (z,c) \implies a_n = (p(z,c),c)$
open whale
$z_{n+1}=h\cdot z_n^2+z_1$
open whale
ah so not k+1
open whale
**cute rizzly bear (won't eat you)**
yeah if you want to formalise it that way
what i prefer formalising it as is
$z_n$ is some algebraic expression in terms of complex numbers and $c$
vivid gulchBOT
yoavmal
open whale
and $p(z)$ is a polynomial of $z$, with coefficients which are polynomials of $c$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
sour sinew
in other words $z_n^2 | z_{n+k} - z_{k+1}$
vivid gulchBOT
cute rizzly bear (won't eat you)
sour sinew
well interesting
open whale
but i have proven it oddly
i think
i want to reprove it
i also think i misphrased it
$z_{n+k}=h\cdot z_n^2+z_k$
vivid gulchBOT
yoavmal
open whale
i think?
sour sinew
ah
vivid gulchBOT
yoavmal
open whale
is this possible?
no, it t is not
but
$z_{n+1}=h\cdot z_n^2+z_1$
that does work
vivid gulchBOT
yoavmal
open whale
so we have a base for thehet induction
now lets try prove it
assume it works for some k
then we want to show it works for k+1
$z_{n+k}=h\cdot z_n^2+z_k$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
$=hz_n^2+z_{k+1}$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
which means that the multiplicities of the roots of $z_n$ are the same in $z_{kn}$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
$z_{n+k}=ahz_n^m+z_k$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
$z_{n+k+1}=a\qty(ahz_n^m+z_k)^m+c$
vivid gulchBOT
yoavmal
open whale
$=a\sum_{j=0}^m\binom{m}{j}(ahz_n^m)^jz_k^{m-j}+c$
vivid gulchBOT
yoavmal
open whale
$=a\sum_{j=1}^m\binom{m}{j}(ahz_n^m)^jz_k^{m-j}+az_k^m+c$
vivid gulchBOT
yoavmal
open whale
$=a\sum_{j=1}^m\binom{m}{j}(ahz_n^m)^jz_k^{m-j}+z_{k+1}$
vivid gulchBOT
yoavmal
open whale
$=a^2hz_n^m\sum_{j=1}^m\binom{m}{j}(ahz_n^m)^{j-1}z_k^{m-j}+z_{k+1}$
vivid gulchBOT
yoavmal
open whale
$=a^2z_n^mhg+z_{k+1}$
vivid gulchBOT
yoavmal
open whale
$=ha^2z_n^m+z_{k+1}$
vivid gulchBOT
yoavmal
open whale
oh so we have even better than previously
i mean, we can insert the a into the h and nothing changes
but i now wonder if we can extract more information from this
$z_{n+2}=a(haz_n^m+c)^m+c$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
$z_{n+k+1}=a^2hz_n^m+z_{k+1}$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
which will entail at least one factoring of a^2 in the final result
yes
$z_{n+k+j}=a^jhz_n^m+z_{k+j}$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
more information is nice
@sour sinew
if you followed the shenanigans
now can we prove anything with induction here
on zkn
$z_{2n}=a^nhz_n^m+z_n$
vivid gulchBOT
yoavmal
open whale
so we have the base case
now we want to do something with zkn and show it for z(k+1)n
well we can always break it down to
$z_{(k+1)n}=a^{nk}hz_n^m+z_{kn}$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
lets check what happens with k+1
$z_{(k+1)n}=z_{kn+n}=a^nhz_{nk}^m+z_n$
so we have that
vivid gulchBOT
yoavmal
open whale
uhh
no, i don't want that
i want it the other way around
$z_{(k+1)n}=a^{nk}hz_{n}^m+z_{nk}$
vivid gulchBOT
yoavmal
open whale
now, with our base induction, we assumed that znk can be written as something with some g
$=a^{nk}hz_n^m+a^ngz_n^m+z_n$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
$=ha^nz_n^m+z_n$
vivid gulchBOT
yoavmal
open whale
right
so, that's what we can obtain
$z_{kn}=ha^nz_n^m+z_n$ is the most informative property we can achieve on this polynomial
vivid gulchBOT
yoavmal
open whale
now, we have not assumed a thing about the connection of a to c
it could be some polynomial of c
but, we do not care
since we can now factor the zn out
and get
$z_{kn}=z_n\qty(ha^nz_n^{m-1}+1)$
vivid gulchBOT
yoavmal
open whale
so this still holds
even if a is a polynomial of degree 1 or more of c
infact
we did not need to assume the +c
we only needed some coefficient
since the recursive relation still holds
in theory c could be some polynomial of another hidden variable
so really we can denote it a0
and a1 being the first coefficient
so what we obtain is
if we explicitly phrase the c in the equation
$z_{kn}=z_n(c)\qty(ha_1(c)^nz_n(c)^{m-1}+1)$
vivid gulchBOT
yoavmal
open whale
i wonder
can we do this for noninteger m
i assume the binomial expansion becomes inconvergent
sour sinew
can we do this for noninteger m
that would definitely be weird
open whale
that would definitely be weird
it would probably break due to the multiplicity of roots
but
maybe it could phrase some riemann surface
which would be increasing in dimension every step
open whale
maybe each z can be multivalued
that would be the result
sour sinew
infinite values
open whale
except if m is irrational
it can still be multivalued
infact
it forms a "multiplicative" group, really additive
if m is irrational it just forms a group isomorphic to the integers
sour sinew
makes sense
open whale
since really
sour sinew
otherwise it's z/qz
vivid gulchBOT
yoavmal
open whale
if m is irrational
what do we get
we get all z^(nm) for integer n
^m
so for each of those we get
$z^{knm^2}$
vivid gulchBOT
yoavmal
open whale
heh, so it is multiplicative
so
it just creates another group isomorphic to the integers
but what about
mapping the whole thing
eh i'm tired
sour sinew
**cute rizzly bear (won't eat you)**
maybe ill try to prove it with my version
open whale
**yoavmal**
anyways @sour sinew what i wanted to show is something like this for a general p(z), not just for z^m+a0 ones
sour sinew
ah
sour sinew
yeah
open whale
ah
well if we've got this proven
the notation is mostly irrelevant
you can phrase it in yours too
now that we have it
sour sinew
true
sour sinew
**yoavmal**
this?
open whale
in terms of our polynomial?
vivid gulchBOT
yoavmal
open whale
but it's not precisely correct
for instance, with what we have right now it's actually
well
for
z^2+c
we get
$z_{n+k}=h\cdot(p(z_n)-c)+z_k$
sour sinew
is that multiplication by h or substitution in h
vivid gulchBOT
yoavmal
sour sinew
hm
open whale
and if we absorb the a^n to the polynomial
for the more recent general form
we get
(i think)
$z_{n+k}=h\cdot(p(z_n)-a_0)+z_k$
vivid gulchBOT
yoavmal
open whale
and this is with 1 term besides a0, a2z^m
lets see how it behaves if we set a1 diffeerent from 0?
like
a2z^m+a1z^k+a0
for k != m
say
k<m
actually
simplest starting point
lets say
z^2+z+c
lets see how it behaves
z0=0
z1=c
z2=c^2+2c
,w expand (c^2+2c)^2+(c^2+2c)+c
sour sinew
wolfram is very helpful
vivid gulchBOT
sour sinew
what is a0
open whale
in mandelbrot we can actually get a convergent sequence of coefficients
open whale
in mandelbrot we can actually get a convergent sequence of coefficients
i.e.
the first n coefficients are the same for the rest of the iterations
sour sinew
or is it from the constant coefficient up
open whale
if we look at z^2+c
we get
0
c
c^2+c
c^4+2c^3+c^2+c
,w expand (c^4+2c^3+c^2+c)^2+c
vivid gulchBOT
open whale
so the first 3 terms are still 2,1,1
and the fourth term, is 5
and will be 5 for the rest
if we were to do another iteration
sour sinew
some power series of z s.t. z^2 + c = z
open whale
some power series of z s.t. z^2 + c = z
well sort of
only the first n coefficients are set
whereas we have 2^n coefficients
which means there is a non-negligent value in each term not deducted by those terms
so if you try to calculate the series with the coefficients it converges to
it would mostly converge to something else
or diverge
etc
also the coefficients converge to the catalan numbers, which is cool
anyways
what else do we get
sour sinew
,w power series of (1-sqrt(1-4c))/2
vivid gulchBOT
open whale
ye but it doesn't work precisely
anyways
we're looking at z^2+z+c now
z0=0
z1=c
z2=c^2+2c
z3=c^4+4c^3+5c^2+3c
,w expand (c^4+4c^3+5c^2+3c)^2+(c^4+4c^3+5c^2+3c)+c
vivid gulchBOT
open whale
z0=0
z1=c
z2=c2+2c
z3=c4+4c3+5c2+3c
z4=c8+8c7+26c6+46c5+50c4+34c3+14c2+4c
is it just me or are the coefficients of the second term the catalan numbers now
sour sinew
it is pretty weird if the catalan numbers start at 4 and drop to 1 7 terms later
open whale
,w expand (14c^2+4c)^2+(14c^2+4c)+c
vivid gulchBOT
open whale
,w catalan numbers
vivid gulchBOT
vivid gulchBOT
open whale
nope, different sequence
sour sinew
,w expand (666c)^2 + (666c) + c
vivid gulchBOT
sour sinew
perfection
open whale
why
sour sinew
idk
open whale
lol
so can we get anything out of this
lets check if we can do some neat algebra to divide multiplicatives
$z_4/z_2=p^2(z_2)/z_2$
vivid gulchBOT
yoavmal
open whale
$\frac{p(z_2)^2+p(z_2)+c}{z_2}$
vivid gulchBOT
yoavmal
open whale
$\frac{(z_2^2+z_2+c)^2+z_2^2+z_2+c+c}{z_2}$
vivid gulchBOT
yoavmal
open whale
uhh wait
why am i doing it like this
oh yeah i can cancel some of the terms already
$\frac{(z_2^2+z_2+c)^2+2c}{z_2}+1+z_2$
vivid gulchBOT
yoavmal
open whale
$\frac{(z_2^4+z_2^2+c^2)+(2z_2^2z_2+2z_2^2c+2z_2c)+2c}{z_2}+1+z_2$
vivid gulchBOT
yoavmal
open whale
$\frac{c^2+2c}{z_2}+(2z_2^2)+z_2^3+z_2+2z_2c+2c+1+z_2$
vivid gulchBOT
yoavmal
open whale
$\frac{z_2}{z_2}+z_2^3+2z_2^2+2z_2c+2c+1+2z_2$
vivid gulchBOT
yoavmal
open whale
$1+z_2^3+2z_2^2+2z_2c+2c+1+2z_2$
vivid gulchBOT
yoavmal
open whale
$z_2^3+2z_2^2+2z_2c+2c+2+2z_2$
vivid gulchBOT
yoavmal
open whale
$z_2^3+2z_2^2+2z_2(c+1)+2c+2$
vivid gulchBOT
yoavmal
open whale
$z_2^3+2z_2^2+2z_2(c+1)+2(c+1)$
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
**yoavmal**
and if we take this writing
we see that this has a nonzero coefficient
that is not a polynomial of z2 nor c
and moreso
we have a free coefficient from z2 as well
meaning that if z2=0
for a particular c
then this whole thing is not 0
can we replicate this result with z6?
$z_6/z_2=\frac{p^4(z_2)}{z_2}$
oo i have an idea
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
vivid gulchBOT
yoavmal
open whale
$\frac{((hz_2)^2+hz_2+c)^2+(hz_2)^2+2c}{z_2}+h$
vivid gulchBOT
yoavmal
open whale
ooo this is perfect
@sour sinew i believe that this form is still correct
like, i can show this is indeed divisible
but the proof here is actually inductive
and general for all multiples of 2
we can probably show this is correct for all znk for all n,k
then we can maybe generalise for a2z^2+a1z+a0
and then maybe to a2z^m+a1z+a0
and then maybe to a2z^m+a1z^k+a0
and then maybe, maybe, to some general polynomial p(z)
given our conditions are sufficient
normal arrow
epic
open whale
also @dusty swallow
while my statement from earlier was incorrect
it was deduced from something that is correct
that being
open whale
**yoavmal**
.
it's just that m can be 1 in theory
but the lemma can be stated as
if zn has a factor r with multiplicity m
then zkn has the same factor r with the same multiplicity m
which works for all a0
even if they're independent of c
vivid gulchBOT
yoavmal
open whale
then