A family of curves in the complex plane may be generated by the formula:
$$ f(t) = e^{it}\left[ 1 - \frac{1}{2}e^{ikt} + \frac{i}{3}e^{-3ikt} \right] $$
for $k=1,2,3,\cdots$. The plotted curve displays $k-$fold rotational symmetry.
Here's the Python code to generate the curve (run as python mystery_curve.py
k
).
This code is also available on my github page.
import sys
import matplotlib.pyplot as plt
import numpy as np
def f(t, k):
"""Return the "Mystery Curve" for parameter k on a grid of t values."""
def P(z):
return 1 - z / 2 - 1 / z**3 / 3j
return np.exp(1j*t) * P(np.exp(k*1j*t))
# k is supplied as a command line argument.
k = int(sys.argv[1])
# Choose a grid of t values at a suitable resolution so that the curve.
# is well-represented.
t = np.linspace(0, 2*np.pi, 200*k+1);
u = f(t, k)
# Plot the Mystery Curve in a pleasing colour, removing the axis clutter.
fig, ax = plt.subplots(facecolor='w')
ax.plot(np.real(u), np.imag(u), lw=2, color='m', alpha=0.5)
ax.set_aspect('equal')
plt.axis('off')
plt.savefig('mystery_curve_{}.png'.format(k))
plt.show()
For example, the curves for $k=3, 6$ and $20$ are plotted below.
Share on Twitter Share on Facebook
Comments
Comments are pre-moderated. Please be patient and your comment will appear soon.
Para 1 year, 4 months ago
Thanks for sharing your knowledge. I see the equation, but I am not able to see how the python code corresponds to it esp the z**3 part of it.
Link | Replychristian 1 year, 2 months ago
Hmm – it's been a while since I wrote this. I think the equation given is just one of a family of curves with different exponent integers: you can experiment. I've made the equation match the code to avoid confusion!
Link | ReplyNew Comment