The Crazy Ant

Consider the plane the crazy ant walks on be a complex plane. The ant starts at the origin $0+0i$ . A turn of $45^\circ$ anticlockwise is equivalent to multiplying by $e^{\frac \pi 4 i}$ . Therefore, the spiral walked by the ant can be expressed as:

$\begin{aligned} Z & = 1 + \frac {e^{\frac \pi 4 i}}2 + \frac {e^{\frac \pi 2 i}}3 + \frac {e^{\frac {3\pi}4 i}}4 + \cdots \\ & = e^{-\frac \pi 4 i} \color{#3D99F6} \left(e^{\frac \pi 4 i} + \frac {(e^{\frac \pi 4 i})^2}2 + \frac {(e^{\frac \pi 4 i})^3}3 + \frac {(e^{\frac \pi 4 i})^4}4 + \cdots \right) & \small \color{#3D99F6} \text{By Maclarin series} \\ & = {\color{#3D99F6} - } e^{-\frac \pi 4 i} \color{#3D99F6} \ln \left(1 - e^{\frac \pi 4 i}\right) \\ & = - e^{-\frac \pi 4 i} \ln \left(1 - \frac 1{\sqrt 2} - \frac i{\sqrt 2} \right) \\ & = - e^{-\frac \pi 4 i} \ln \left(\sqrt{2-\sqrt 2} \cdot e^{-i \tan^{-1}(\sqrt 2+1)} \right) \\ & = - e^{-\frac \pi 4 i} \left(\frac {\ln (2-\sqrt 2)}2 -i \tan^{-1}(\sqrt 2+1) \right) \\ & = e^{-\frac \pi 4 i} \left(\sqrt{\frac {\ln^2 (2-\sqrt 2)}4 + (\tan^{-1}(\sqrt 2+1))^2} e^{-i\tan^{-1} \frac {2\tan^{-1}(\sqrt 2+1)}{\ln (2-\sqrt 2)}}\right) \end{aligned}$

The distance between the ant's final and initial positions is given by $|Z| = \sqrt{\frac 14 \ln^2 (2-\sqrt 2) + (\tan^{-1}(\sqrt 2+1))^2} \approx \boxed{1.208}$ .

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Reference: Maclaurin series

The displacements in the E-direction are $1-\frac{1}{5}+\frac{1}{9}- \frac{1}{13}+... = 4 \sum_{n=0}^{\infty} {\frac{1}{(8n+1)(8n+5)}}=\frac{π+\ln(3+2\sqrt 2)}{4 \sqrt{2}} = 0.86697298...=A$

The displacements in the NE-direction are $\frac{1}{2}-\frac{1}{6}+\frac{1}{10}-\frac{1}{14}+... = 4 \sum_{n=0}^{\infty} \frac{1}{(8n+2)(8n+6)}=\frac{π}{8} = 0.39269908...=B$

The displacements in the N-direction are $\frac{1}{3}-\frac{1}{7}+\frac{1}{11}-\frac{1}{15}+... = 4 \sum_{n=0}^{\infty} \frac{1}{(8n+3)(8n+7)}=\frac{π - 2 \ln(1+ \sqrt{ 2})}{4 \sqrt{2}}= 0.24374774...=C$

The displacements in the NW-direction are $\frac{1}{4}-\frac{1}{8}+\frac{1}{12}-\frac{1}{16}+... = 4 \sum_{n=0}^{\infty} \frac{1}{(8n+4)(8n+8)}=\frac{\ln(2)}{4} = 0.17328679...=D$

The total net displacement in the positive x-direction is $\Delta x= A+\frac{1}{2}\sqrt{2}(B-D)=1.02212141..$

The total net displacement in the positive y-direction is $\Delta y= C+\frac{1}{2}\sqrt{2}(B+D)=0.64396018..$

The overall net displacement is $\sqrt{(\Delta x)^2+ (\Delta y)^2}=\boxed{1.20806328...}$

Fourier series can also be used once you get the distance interms of sinnx/n series ,cosnx/n series.