# Infinite staircase to a circle

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.

• 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,...? Oct 10, 2014 at 10:46
• That's another problem I guess Oct 10, 2014 at 10:48
• 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.
– Paul
Oct 10, 2014 at 13:48
• 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. Oct 11, 2014 at 17:52
• The first digits of $x_{\infty}$ are $0.7808861196194307100503584709816329393433\dotsc$ and this does not appear to be an algebraic number. Hope this helps Oct 12, 2014 at 15:43

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

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

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:

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}]

• 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. Oct 13, 2014 at 6:17
• (I'll award the bounty when it becomes possible to do so.) Oct 13, 2014 at 6:25
• I don't see how this answers the question...? Oct 13, 2014 at 8:09
• 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 $1/\sqrt{2}$. Oct 13, 2014 at 9:37
• @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 ... Oct 13, 2014 at 9:52