Infinite staircase to a circle

Solution 1:

Not a nice, clean answer I'm afraid - just some observations.

Your staircase looks like this:

As noted below,

$$x_{n}=x_{n-1}+\dfrac{1}{2}\left(\sqrt{1-y_{n-1}\ ^{2}}-x_{n-1}\right)\\ y_{n}=y_{n-1}+\dfrac{1}{2}\left(\sqrt{1-x_{n}\ ^{2}}-y_{n-1}\right)$$

with first few terms

\begin{align} x_0&\quad\dfrac{1}{2}\\ y_0&\quad\dfrac{\sqrt{3}}{4}\\ x_1&\quad\frac{1}{8} \left(2+\sqrt{13}\right)\\ y_1&\quad\frac{1}{16} \left(\sqrt{47-4 \sqrt{13}}+2 \sqrt{3}\right)\\ x_2&\quad\frac{1}{32} \left(\sqrt{-4 \sqrt{3 \left(47-4 \sqrt{13}\right)}+4 \sqrt{13}+197}+2 \sqrt{13}+4\right)\\ \dots \end{align}

It gets a bit rediculous after that, so decimal approximation is preferable, as given in the comments.

We start to get a fuller picture though by taking steps of $\dfrac{1}{k}$ instead of $\dfrac{1}{2}$, so our sequence becomes

$$x_{n}=x_{n-1}+\dfrac{1}{k}\left(\sqrt{1-y_{n-1}\ ^{2}}-x_{n-1}\right)\\ y_{n}=y_{n-1}+\dfrac{1}{k}\left(\sqrt{1-x_{n}\ ^{2}}-y_{n-1}\right)$$

with first few terms:

\begin{align} x_0&\quad\dfrac{1}{k}\\ y_0&\quad\dfrac{\sqrt{1-\dfrac{1}{k^2}}}{k}\\ x_1&\quad\dfrac{\sqrt{\dfrac{1}{k^4}-\dfrac{1}{k^2}+1} k+k-1}{k^2}\\ y_1&\quad\dfrac{\sqrt{1-\dfrac{1}{k^2}} k-\sqrt{1-\dfrac{1}{k^2}}+\sqrt{1-\dfrac{\left(\sqrt{\dfrac{1}{k^4}-\dfrac{1}{k^2}+1} k+k-1\right)^2}{k^4}} k}{k^2}\\ \dots \end{align}

which looks like this:

where clearly $\lim_{k\rightarrow\infty}\arctan\dfrac{x_{n}}{y_{n}}=\dfrac{\pi}{4}$, and $\lim_{k\rightarrow\infty}\{x_{n},y_{n}\}=\{\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\}.$

Note

Fairly good approximation for point on $k$ steps is

$$\left\{\frac{e^{2/\pi }}{\sqrt{\left(\left(\frac{1}{k-\zeta (3)}+1\right)^{k-\zeta (3)}\right)^{4/\pi }+e^{4/\pi }}},\frac{\left(\left(\frac{1}{k-\zeta (3)}+1\right)^{k-\zeta (3)}\right)^{2/\pi }}{\sqrt{\left(\left(\frac{1}{k-\zeta (3)}+1\right)^{k-\zeta (3)}\right)^{4/\pi }+e^{4/\pi }}}\right\}$$

Manipulate[range = 100;
 h[{x_, y_}] := {x + 1/k (Sqrt[1 - y^2] - x), 
y + 1/k (Sqrt[1 - (x + 1/k (Sqrt[1 - y^2] - x))^2] - y)};
 x0 = N[0 + 1/k (Sqrt[1 - 0^2] - 0)];
 y0 = N[0 + 1/k (Sqrt[1 - x0^2] - 0)];
 nl = NestList[h, {x0, y0}, range];
 Show[Graphics[{Circle[{0, 0}, 1],
Join[{{0, 0}, {1/k, 0}}, 
  Flatten[{{nl[[#, 1]], nl[[#, 2]]}, {nl[[# + 1, 1]], 
       nl[[#, 2]]}} & /@ Range[range], 1]] // Line,
Join[{{0, 0}}, 
  Flatten[{{nl[[#, 1]], nl[[#, 2]]}} & /@ Range[range], 1]] // 
 Line,
Join[{{1/k, 0}}, 
  Flatten[{{nl[[# + 1, 1]], nl[[#, 2]]}} & /@ Range[range], 1]] //
  Line, {{0, 0}, {1/Sqrt[2], 1/Sqrt[2]}} // Line,
Red, PointSize[Large], 
Point[{p1 = {(k - k^3 + 2*Sqrt[1 + k^(-4) - k^(-2)]*k^4 + 2*k^5 + 
       Sqrt[2]*
        Sqrt[(-1 + 3*k^2 - Sqrt[1 + k^(-4) - k^(-2)]*k^3 - 
            3*k^4 + 2*k^6)*(-1 + 4*k^2 - 3*k^4 - 
            2*Sqrt[1 + k^(-4) - k^(-2)]*k^5 + 2*k^6)])/(-1 + 
       5*k^2 - 4*k^4 + 
       4*k^6), (Sqrt[
        1 - k^(-2)]*(1 - 4*k^2 + 3*k^4 + 
         2*Sqrt[1 + k^(-4) - k^(-2)]*k^5 - 2*k^6 + 
         Sqrt[2]*k*
          Sqrt[(-1 + 3*k^2 - Sqrt[1 + k^(-4) - k^(-2)]*k^3 - 
              3*k^4 + 2*k^6)*(-1 + 4*k^2 - 3*k^4 - 
              2*Sqrt[1 + k^(-4) - k^(-2)]*k^5 + 2*k^6)]))/((-1 + 
         Sqrt[1 + k^(-4) - k^(-2)]*k)*(-1 + 5*k^2 - 4*k^4 + 
         4*k^6))},
  {1/Sqrt[2 - k^(-2)], Sqrt[(-1 + k^2)/(-1 + 2*k^2)]}
  }],
{{1/k, 0}, p1} // Line,
Line[{{0, 0}, {Cos[ArcTan[y0/x0]], Sin[ArcTan[y0/x0]]}}],
Blue, PointSize[Large], 
Point[{E^(2/Pi)/
   Sqrt[E^(4/Pi) + ((1 + (k - Zeta[3])^(-1))^(k - Zeta[3]))^(4/
        Pi)], ((1 + (k - Zeta[3])^(-1))^(k - Zeta[3]))^(2/Pi)/
   Sqrt[E^(4/Pi) + ((1 + (k - Zeta[3])^(-1))^(k - Zeta[3]))^(4/
        Pi)]}]
}, PlotRange -> {{0, 1}, {0, 1}}, Frame -> True, 
ImageSize -> 400](*,Plot[(((1+1/(k-Zeta[3]))^(k-Zeta[3]))/E)^(2/
Pi)x,{x,-1,1}]*)], {{k, E}, 1.3, 30, 0.01}]