# and e

Mathematics and other neat things

This video shows continuous variation in angle (from 0 to Pi) of the connect-the-dots algorithm while keeping all other parameters fixed. Each instant is a valid 2-coloring.

Music: “Derbyshire” by Nerve

Mathematica code:

Manipulate[  ImageCrop[   Graphics[    GraphicsComplex[     Table[      {-(.96)^n*Sin[n*.001 f], .96^n*Cos[n*.001 f]},      {n, 0, 300}],    Polygon[Table[i, {i, 1, 300, 1}]]],   PlotRange -> .1, ImageSize -> 640], {640, 480}],{f, 1, 3145, 1}]

Let $$S_n (z) = \sum_{k=0}^n z^k$$ then parametrize $$S_n$$ on circles in the complex plane centered at the origin. The animation shows the graph of the real part of $$S_{100} (r e^{2\pi i t})$$ as a function of $$t$$ and how it changes as $$r$$ ranges from $$0$$ to $$1$$.

### Source Code

Source code for the previous post. Not quite complete since I ended up using Unix commands to put the individual pictures together.

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.colors as mplcolors

width = 256
height = 256
length = 40
maxiter = 40

fig = plt.figure(figsize=(float(width)/100, float(height)/100), edgecolor=’k’)
fig.subplots_adjust(left=0.0, right=1.0, top=1.0, bottom=0.0, wspace=0.0, hspace=0.0)

x1 = -1.0
x2 = 1.0

y1 = (x1 - x2)/2 * float(height)/width
y2 = (x2 - x1)/2 * float(height)/width

y, x = np.ogrid[y1:y2:height*1j, x1:x2:width*1j]

cdict = {‘red’: ((0.0, 0.4, 0.4),
(0.5, 0.7, 0.7),
(1.0, 1.0, 1.0)),
‘green’: ((0.0, 0.0, 0.0),
(0.5, 1.0, 1.0),
(1.0, 1.0, 1.0)),
‘blue’: ((0.0, 1.0, 1.0),
(0.5, 1.0, 1.0),
(1.0, 1.0, 1.0))}
my_cmap = mplcolors.LinearSegmentedColormap(‘my_colormap’, cdict, 256)

for n in xrange(length):
t = float(n+1)/length
e = np.exp(2*np.pi*t*1j)
c = e/2*(1-e/2)

z = x + y*1j
divtime = maxiter + np.zeros(z.shape, dtype=int)

for k in xrange(maxiter):
z = z*z + c
diverge = z*np.conj(z) > 4
div_now = diverge & (divtime == maxiter)
divtime[div_now] = k
z[diverge] = 2

ax.cla()
ax.imshow(divtime, cmap=my_cmap)
ax.axis(‘off’)
filename = ‘JuliaSet_{:03}.png’.format(n+1)
fig.savefig(filename)

Julia set for $$f_c(z) = z^2 + c$$ with $$c = \frac{e^{2 \pi i t}}{2} (1 - \frac{e^{2 \pi i t}}{2} )$$ where $$t \in [0,1]$$. These values of $$c$$ correspond to the boundary of the main cardioid of the Mandelbrot set.

Various values of $$1^\pi$$ in the complex plane. It is worth noting that the seventh value is very close to $$1$$. This corresponds to $$\frac{22}{7}$$ being a very good rational approximation to $$\pi$$. Also very interesting: the next closest to $$1$$ is the $$106^{th}$$ value. This corresponds to the second convergent, $$\frac{333}{106}$$, in the continued fraction expression for $$\pi$$.

### Basel Problem

First we want to an expression for
$f(z) = \sum_{n=1}^\infty \frac{1}{n}z^n.$
We do this by noting that
$f’(z) = \sum_{n=0}^\infty z^n = \frac{1}{1-z}.$
So
$f(z) = \int \frac{1}{1-z} \mathrm{d}z = -\log (1-z) + C.$
Evaluating both sides at $$z=0$$ shows that $$C=0$$. Now let $$z=e^{ix}$$ and equate imaginary parts
$\sum_{n=1}^\infty \frac{\sin (n x)}{n} = -\mathfrak{Im} (\log(1-e^{ix})).$
However, for any $$w\in\mathbb{C}$$, $$\log(w) = \log(|w|e^{i\mathrm{Arg}(w)}) = \log(|w|) + i\mathrm{Arg}(w)$$ and $$\mathrm{Arg}(w) = \arctan (\frac{\mathfrak{Im}(w)}{\mathfrak{Re}(w)})$$. Now, we have $$1-e^{ix} = 1-\cos(x) + i\sin(x)$$ so $$\mathrm{Arg}(1-e^{ix}) = \arctan (\frac{\sin(x)}{1-\cos(x)}) = \arctan(\frac{2\sin(\frac{1}{2}x)\cos(\frac{1}{2}x)}{2\sin^2(\frac{1}{2}x)}) = \arctan(\cot(\frac{1}{2}x))$$, which is equal to $$\frac{\pi}{2} - \frac{1}{2}x$$. So finally we get
$\sum_{n=1}^\infty \frac{\sin(nx)}{n} = \frac{1}{2}x - \frac{\pi}{2}.$
We can now find the anti-derivative of both sides to get
$-\sum_{n=1}^\infty \frac{\cos(nx)}{n^2} = \frac{1}{4}x^2 - \frac{\pi}{2}x + C.$
But we can find $$C$$ by noting that
$-\int_0^{2\pi} \sum_{n=1}^\infty \frac{\cos(n x)}{n^2} \mathrm{d}x = 0.$
We have
$\int_0^{2\pi} \frac{1}{4}x^2 - \frac{\pi}{2}x + C \;\mathrm{d}x = [\frac{1}{12}x^3 - \frac{\pi}{4}x^2 + Cx]^{2\pi}_0 = \frac{\pi^3}{3} + 2\pi C = 0.$
So $$C = -\frac{\pi^2}{6}$$ and we have established the equality
$-\sum_{n=1}^\infty \frac{\cos(nx)}{n^2} = \frac{1}{4}x^2 - \frac{\pi}{2}x - \frac{\pi^2}{6}.$
If we let $$x=0$$ we see that
$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$

A curve in $$\mathbb{C}$$ given by $$\zeta(\frac{1}{2}+it)$$. Here $$\zeta(s)$$ refers to the analytic continuation of
$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}.$

### Mathjax

I just put mathjax into my theme, so I am testing it out. If p is an odd prime, then