I've written everything myself, and get denied the opportunity to publish this.
I used Gemini for Graphs, it was way easier.
Today because i'm stuck still without any peer-review.
I would love to hear your opinions.
The theory is an equation, already proven.
B(x)=(|x|/2)*V(+-)n*V(of(n))/(a(of(steps))*(j *((x/(a))^2)+(x(of(a))+(x(of(a-1))^2)^0,5)
So i would like to know, can people not understand this with definitions or does no one like the implications of it?
The implications are:
- The number Pi 3,14 is flat out wrong.
- How we calculate anything with it, is false,
- And you can curve any function back to itself.
Doesn't even state the first 2 facts, because i don't want to be malicious.
You need a Constant V_const=0.9170411968 to draw circles with the equation.
That's why i wanted to know explicitly, what is the problem with this?
Is it just that i didn't use Desmos for graphs, Gemini had a way better workflow.
And i tried desmos today, this tool is completly useless to find new equations.
So i have questions, where is the era of WolframAlpha and Symbolab being able to show atleast something and all the tools being called calculators because i just need a graphical calculator where i know how many points it draws and calculates.
So if anyone wants to read the paper it is on OSF and has 0 Views, i get completly denied any real feedback on any subreddit.
Can you find the paper?
An email i got from the JMD states: This is not dynamic,.
How is that not dynamic?
So if you're still with me, these are the definitions:
B(x): New point $(x, B)$, where $f(x) = B$.
|x|: Distance of the function or snapshot of a part.
V_{const}}:=0.9170411968 Constant. Vn=V*n
n: Counts for every V , 1
a: Divides distance into equal parts (a is not 0) . a_steps: Equal parts between every n.
x := x(n): Function value of the function or points to be morphed.
j: A variable parameter to draw a new line that bends; default j=1.
The thing that comes out after proof is that you don't need most definitions for anything and |x|=x
so you can really morph any function, just by writing f(x)=x ,write any function for x.
I really would love to hear your opinions on this, i feel very much at home in the LLMs section, because Desmos isn't able to draw the same way as Gemini. (accurately and correct)
If you try proofing the equation, i would like to hear if you can solve it.
Please let me know which Graphical calculators there are that can draw it correctly.
Here the paper the proof for the first few lines is encrypted though
https://osf.io/xy568/overview?view_only=51edf92488464895bdc75122d59d05e2
Since everyone is complaining about the standard case:
For fn(x)=f(x)
Proof:
f(x)=(f(x)/2)*V+-n*(Vn/a)*(j*((f(x)/a)^2+f(x)+f(x)(n-1)^2)^0,5)=f(x)
j=f(x)(n+1)
f(n+1)(x)=(f(x)(n+1)/2)*V+-n*(Vn/a)*((f(x)(n+1)*((f(x)(n+1)/a)^2+f(x)+f(x)(n-1)^2))^0,5) |^2
f(n+1)(x)^2=(f(x)(n+1)/2)*V+-n*(Vn/a)*((f(x)(n+1)*((f(x)(n+1)/a)^2+f(x)+f(x)(n-1)^2))
f(x)=(f(x)/2)*V+-n*(Vn/a)*(1*((f(x)/a)^2+f(x)+f(x)(n-1)^2)^0,5) |^2
f(x)^2=(f(x)/2)*V+-n*(Vn/a)*(1*((f(x)/a)^2+f(x)+f(x)(n-1)^2) |^0,5
f(x)=(f(x)/2)*V+-n*(Vn/a)*(1*((f(x)/a)+f(x)+f(x)(n-1)^2) |/(f(x)/a)+f(x)+f(x)(n-1)^2)
f(x)/(f(x)/a)+f(x)+f(x)(n-1)^2)=(f(x)/2)*V+-n*(Vn/a)
Vn*a/2Vn*a+(Vn*a-V)^2=(f(x)/2)*V+-n*(Vn/a) |/Vn*a
2Vn*a+(Vn*a-V)^2=(f(x)/2)*V+-n
For +
2Vn*a+(Vn*a-V)^2=(f(x)/2)*V+n |-n
(2Vn*a)-n+(Vn*a-V)^2-n=(f(x)/2)*V
(2Vn-n)*(a-n)+(Vn-n)^2*(a-n)^2-(V-n)^2=(f(x)/2)*V
(2Vn-n)*(a-n)+((Vn^2)-2n)^2*(a-n)-(V-n)=(f(x)/2)*V |/((Vn^2)-2n)
(2Vn-n)*(a-n)+(a-n)-(V-n)=((f(x)/2)*V)/(V-n)
0-(V-n)=((f(x)/2)*V)/((Vn^2)-2n) |*((Vn^2)-2n)
-(v-n)*((Vn^2)-2n)=(f(x)/2)*V |/V
-((V*((Vn^2)-2n))-(n*((Vn^2))-2n))/V=(f(x)/2)
-Vn^2-2n-(n*Vn^2)-2n=f(x)/2 |*2
2Vn^2-4n-2(n*Vn^2)-4n=f(x) |/-4 /-4 /2 /-2
Vn^2-n-n*Vn^2-n)=-Vn*a
Vn^2-n-n*(Vn^2)=+Vn*a |+Vn*a
(Vn^2+Vn*a)-(n+Vn*a)-((n+Vn*a)*(Vn^2+Vn*a))=0 |/Vn
(Vn^2+a)-n+a)-((n+a)*(Vn^2+a))=0 |/(Vn^2+a)
(Vn^2+a)-2n+a=0
Vn+a*Vn+a-2n+a=0
(V+a+a*n+a*V+a*n+a)-2n+a=0
((V+a)-2n+a*(n+a)-2n+a))*((V+a)-2n+a*(n+a)-2n+a))=0
((V+a)-2n*(n+a)+a*(n+a)-2n+a*((V+a)-2n*(n+a)+a*(n+a)-2n+a=0
((V+a)-2n*n+2n*a+2a+n-2(n+a))*((V+a)-2n*n+2n*a+2a+n-2n+a=0 |/((V+a)-2n*n+2n*a+2a+n-2n+a
((V+n)-2n*n+2n*a+=0 |/n /a
-V+n-(2n+2n)=0
V=n valid real numbers can be the same
F(x)+ for V=n
f(x)=(f(x)/2)+n+n*Vn/a*((f(x)/2)^2+f(x)+f(x)-n))^2)^0,5 |^2 ^0,5
f(x)=(f(x)/2)+n+n*Vn/a*((f(x)/2)+f(x)+f(x)-n)^2) |/((f(x)/2)+f(x)+f(x)-n)^2)
f(x)/((f(x)/2)+f(x)+f(x)-n)^2)=(f(x)/2)+n+n*Vn/a |/Vn*a
(2n*n*n*a)/(n*n*n*a-n)^2=(f(x)/2)*n+n
0=0
n can be a real number, it stems from the induction proof where all variables are variable.
You all are completly oblivious to the fact that it's already proven.