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Suppose you start at $(0,0)$ on the unit disc and repeat the following procedure again and again:

  1. Face east and walk half-way to the circumference.
  2. Face north and walk half-way to the circumference.

What is your limiting position $(x,y)$?

This is a fun problem thought of by a friend. I'm interested to see if anyone can find a nice, clean solution to it.

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  • $\begingroup$ I assume you keep doing steps of the type described. But do you alternate east, north, east, north,...; or is the pattern of directions east, north, west, south, east, north, west, south,...? $\endgroup$
    – paw88789
    Oct 10, 2014 at 10:46
  • $\begingroup$ That's another problem I guess $\endgroup$ Oct 10, 2014 at 10:48
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    $\begingroup$ If $x_n$ and $y_n$ is the position after one north east then north move then $$x_{n+1}=x_n+\frac{1}{2}((1-y_n^2)^{\frac{1}{2}}-x_n)= \frac{1}{2}(x_n+(1-y_n^2)^{\frac{1}{2}})$$ $$y_{n+1}=y_n+\frac{1}{2}((1-x_{n+1}^2)^{\frac{1}{2}}-y_n)= \frac{1}{2}(y_n+(1-x_{n+1}^2)^{\frac{1}{2}})$$ We know the limits of $x_n$ and $y_n$ exist. Letting them be x and y and substituting into the equations as limit values does not let us find them though. We simply get that x and y lie on the circle. $\endgroup$
    – Paul
    Oct 10, 2014 at 13:48
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    $\begingroup$ The points on the unit disc for which the procedure (starting with an eastward move) produces the same limiting position lie on a continuous curve passing through (0,-1), (0,0) and the limiting position. Points on another curve, passing through (0,-1), ($\tfrac{1}{2},\tfrac{1}{2}$) and the limiting position, have the same property if we start with a northward move. Some progress might be made by attempting to describe these curves. $\endgroup$
    – MartinG
    Oct 11, 2014 at 17:52
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    $\begingroup$ The first digits of $x_{\infty}$ are $0.7808861196194307100503584709816329393433\dotsc$ and this does not appear to be an algebraic number. Hope this helps $\endgroup$ Oct 12, 2014 at 15:43

1 Answer 1

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

Your staircase looks like this:

enter image description here

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:

enter image description here

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}]
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  • $\begingroup$ This is a great answer, thanks. I guess heuristically it explains why we get so close to $\sqrt{2} = 2 \cdot \frac{1}{\sqrt{2}}$ when adding up the co-ordinates for the $k = 2$ case. $\endgroup$
    – Mr. Chip
    Oct 13, 2014 at 6:17
  • $\begingroup$ (I'll award the bounty when it becomes possible to do so.) $\endgroup$
    – Mr. Chip
    Oct 13, 2014 at 6:25
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    $\begingroup$ I don't see how this answers the question...? $\endgroup$ Oct 13, 2014 at 8:09
  • $\begingroup$ I'm satisfied that $x_\infty$ is not a remarkable number with a simple closed form. Of course I'd welcome more information on its properties, and maybe the previous sentence is actually in error (despite my checks at the OEIS), but I think viewing things from the perspective of $1/k$ suggests it's just "some number" in a certain sequence converging to $1/\sqrt{2}$. $\endgroup$
    – Mr. Chip
    Oct 13, 2014 at 9:37
  • $\begingroup$ @Joshua Ciappara Glad you like it. Feel free to change accept though if you get a better answer :) It is an interesting problem! Expanding the terms doesn't look hopeful for a closed solution - but you never know ... $\endgroup$
    – martin
    Oct 13, 2014 at 9:52

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