Pizza and Problems/Pizza and Problems 1 (7/28/07)

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Use this page to post solutions to our first Pizza and Problems session, held 7/28/07. Examine the original problem set.

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Student Solutions to the Problems in Pizza and Problems Set #1

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Problem 7: Solution by David Arnold

The following image shows the general idea of the problem.

Starting at the origin, we walk one unit to the right. Then we make a left turn and walk half the previous distance. Another left turn, then walk the half the previous distance, etc. My thought was that the geometry of complex numbers and vectors would be a good solution. I thought of the individual pieces of the walk as complex numbers, pictured as vectors in the complex plane in the following image.

Recall the following fact:

$r\,e^{i\theta}=r\,(\cos\theta+i\sin\theta)$

Here, $r$ is the magnitude of the complex number (or vector) and $θ$ is the argument of the complex number (the angle made with the positive x-axis). Thus, the four vectors pictured above can be written in sequence:

$1e^{i0},\ \frac12 e^{i\pi/2},\ \frac14e^{i\pi},\ \frac18e^{i3\pi/2},\ \dots$

Thus, to find the final arrival point, we would sum the vectors as follows:

$\sum_{k=0}^{\infty} \frac{1}{2^k}\,e^{ik\pi/2} =e^{i0}+\frac12 e^{i\pi/2}+\frac14 e^{i\pi}+\frac18 e^{i3\pi/2}+\cdots$

Now, this infinite series has the form of a geometric series

$S=a+ar+ar^2+ar^3+\cdots,$

which has the sum

$S=\frac{a}{1-r},$

provided $| r | < 1$ . In our case, the series

$S=\sum_{k=0}^{\infty} \frac{1}{2^k}\,e^{ik\pi/2} =\sum_{k=0}^{\infty} \left[\frac{1}{2}\,e^{i\pi/2}\right]^k$

has the form

$\sum_{k=0}^\infty ar^k$

where $a = 1$ and

$r=\frac12e^{i\pi/2}.$

Moreover, note that the magnitude of $r=\frac12e^{i\pi/2}$ (its length) is 1/2. Thus, the series is geometric and its sum is given by $a / (1 - r)$ , or