Topic: What's the closed-form expression for y=f(x) describing a cycloid curve?

+Anonymous A — 10 months ago #66,434

https://desmos.com/calculator/kzjfgtozep

·Anonymous A (OP) — 10 months ago, 8 minutes later[T] [B] #661,958

(t-sin(t),1-cos(t))

·Anonymous A (OP) — 10 months ago, 24 minutes later, 32 minutes after the original post[T] [B] #661,963

You could also use

((6**(2/3))/2)*x**(2/3)

to fix the curve starting at the initial cusp (zoom in at the origin (0,0) until you'll see the error), just figuring out different ways to approximate.

+Anonymous B — 10 months ago, 12 minutes later, 45 minutes after the original post[T] [B] #661,967

(x+y)^50-(x-y)^50=x^2-y^2

That should do the trick.

·Anonymous B — 10 months ago, 1 minute later, 46 minutes after the original post[T] [B] #661,968

@previous (B)
I studied white math.

·Anonymous A (OP) — 10 months ago, 18 minutes later, 1 hour after the original post[T] [B] #661,971

@661,963 (A)
This problem is also related to Kepler's equation…

function E(e,M,n){
    let E=M;
    for(let k=1;k<=n;k++)
        E=M+e*Math.sin(E);
    return E;
}
//(E,e,M)=>E+(((M+e*Math.sin(E)-E)*(1+e*Math.cos(E)))/(1-(e*e*(Math.cos(E)**2))));

(Edited 23 seconds later.)

+Anonymous C — 10 months ago, 4 minutes later, 1 hour after the original post[T] [B] #661,972

Valid JavaScript code be like:

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·Anonymous A (OP) — 10 months ago, 7 minutes later, 1 hour after the original post[T] [B] #661,973

@previous (C)
At least put the JSFuck inside [code=javascript][/code], it's too big…

·Anonymous C — 10 months ago, 1 minute later, 1 hour after the original post[T] [B] #661,974

@previous (A)
> At least put the JSFuck inside

, it's too big…

It’s almost as big as…

·Anonymous A (OP) — 10 months ago, 25 minutes later, 1 hour after the original post[T] [B] #661,985

I wonder if I can still post despite the fatal errors.

·Anonymous A (OP) — 10 months ago, 12 hours later, 13 hours after the original post[T] [B] #662,013

I wonder how you'd write iterative functions in Desmos without doing f(f(f(f(x))))…

+Anonymous D — 10 months ago, 11 minutes later, 13 hours after the original post[T] [B] #662,014

@previous (A)
Always remember to end the equation by dividing by Zero.
A positive or negative number when divided by zero is a fraction with the zero as denominator. Zero divided by a negative or positive number is either zero or is expressed as a fraction with zero as numerator and the finite quantity as denominator. Zero divided by zero is zero.

·Anonymous A (OP) — 10 months ago, 33 minutes later, 14 hours after the original post[T] [B] #662,015

@661,971 (A)
@662,013 (A)
Here's another attempt at approximating (still not good, especially at the cusps, and it's kinda slow, there are redundant variables in the inverse Kepler formula): https://desmos.com/calculator/uygh3pmg8u
There are more methods.

·Anonymous A (OP) — 10 months ago, 1 hour later, 16 hours after the original post[T] [B] #662,023

Area under cycloid (3πr²): https://desmos.com/calculator/iaxuo25pen
(I don't know what I'm doing)

·Anonymous A (OP) — 10 months ago, 35 minutes later, 16 hours after the original post[T] [B] #662,028

The fact that there's no simple elegant solution as of yet feels like that annoying moment when you've gotta sneeze but you can't and it just won't happen and you're trapped in suspense, teetering on the edge, in a cycle of despair being so desperate, craving for that sweet satisfying great relief. (I'm sorry for writing this more detailed than necessary, umm…)

·Anonymous A (OP) — 10 months ago, 1 hour later, 18 hours after the original post[T] [B] #662,033

@662,015 (A)
Using Newton's method and Taylor series to get the inverse of x-sin(x): https://desmos.com/calculator/ngefjqgfnr

·Anonymous A (OP) — 9 months ago, 1 month later, 1 month after the original post[T] [B] #667,295

i(x)=lerp(lerp(cbrt(6*x),TAU+cbrt(6*(x-TAU)),x/TAU),PI+2*arcsin((x/PI)-1),sin(x/2)*.5);
f(x,n)={n=0:i(x),x+sin(f(x,n-1))};
s(x)=((f(mod(x,TAU),N)+f(mod(x,TAU),N+1))/2)+TAU*floor(x/TAU);

https://desmos.com/calculator/zmpztnlror
Still thinking about this as of today… I might as well even look more into Chebyshev interpolation and asymptotic series expansion…

+Anonymous E — 9 months ago, 21 minutes later, 1 month after the original post[T] [B] #667,299

@previous (A)
lol

Lerp.

·Anonymous A (OP) — 9 months ago, 9 hours later, 1 month after the original post[T] [B] #667,339

function inverse(y,x0=y+Math.sin(y+Math.sin(y+Math.sin(y))),eps=1e-3,maxIter=100){
    let xPrev=x0,
        xCurr=x0+1e-4; //Initial small perturbation to avoid division by zero
    for(let i=0;i<maxIter;i++){
        const fCurr=xCurr-Math.sin(xCurr)-y;
        if(Math.abs(1-Math.cos(xCurr))>=eps){
            //Newton-Raphson step
            const df=1-Math.cos(xCurr);
            let delta=fCurr/df,
                alpha=1;
            while(Math.abs(delta*alpha)>1e-2)
                alpha*=.5;
            xNext=xCurr-delta*alpha;
        }else{
            //Secant step with safeguards
            const fPrev=xPrev-Math.sin(xPrev)-y;
            if(Math.abs(fCurr-fPrev)<1e-12){
                //Avoid division by zero - take small step
                xNext=xCurr-1e-4;
            }else{
                const delta=(fCurr*(xCurr-xPrev))/(fCurr-fPrev);
                let alpha=1;
                while(Math.abs(delta*alpha)>1e-2)
                    alpha*=.5;
                xNext=xCurr-delta*alpha;
            }
        }
        xPrev=xCurr;xCurr=xNext;
        if(Math.abs(fCurr)<1e-6)break;
    }
    return xCurr;
}

THIS SHIT STILL SUCKS, IT NEEDS TOO MANY ITERATIONS

+Anonymous F — 9 months ago, 2 days later, 1 month after the original post[T] [B] #667,761

The third-order Householder's method works perfect: https://desmos.com/calculator/azs61txtm9

·Anonymous A (OP) — 9 months ago, 12 hours later, 1 month after the original post[T] [B] #667,901

@previous (F)
f(x) = x-sin(x)
d/dx f^-1(x) = 1/(f'(f^-1(x))) = 1/(1-cos(f^-1(x)))
1/(d/dx f^-1(x)) = 1-cos(f^-1(x))

Make of it what you will…

(Edited 2 minutes later.)

TinyChan