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Keybase proof

I hereby claim:

  • I am jackolantern on github.
  • I am jackolantern (https://keybase.io/jackolantern) on keybase.
  • I have a public key whose fingerprint is 43EF C369 8F77 6EA8 2612 C720 38C9 111F 103D 8FDA

To claim this, I am signing this object:

Fungal Overgrowth
Peeling layers
Relaxation
Stress
wash your hands
Geometric symmetry
Strange physics / strange topologies
the enemy is on the inside
postcards
everything almost fits... almost
@jackolantern
jackolantern / springer-free-maths-books.md
Created December 29, 2015 02:58 — forked from bishboria/springer-free-maths-books.md
Springer have made a bunch of maths books available for free, here are the direct links
@jackolantern
jackolantern / rearrange_power_of_three.py
Last active December 16, 2015 03:51 — forked from Radcliffe/rearrange_power_of_three.py
Powers of 3 having the same digits
"""
James Tanton @jamestanton asked:
Is there a power of three whose digits can be rearranged to form
another power of three?
(https://twitter.com/jamestanton/status/676773021223272448)
The following Python script searches for examples. There is no power
of three, less than 3^200000, whose digits in base 10 can be rearranged
to form another power of three. In base 4 there is no number less than
3^29 with this property.
"""
### Keybase proof
I hereby claim:
* I am jackolantern on github.
* I am jackolantern (https://keybase.io/jackolantern) on keybase.
* I have a public key whose fingerprint is 03E4 EF54 F32E B0B7 8591 CBAB 757D E439 B135 C3D3
To claim this, I am signing this object:

Keybase proof

I hereby claim:

  • I am jackolantern on github.
  • I am jackolantern (https://keybase.io/jackolantern) on keybase.
  • I have a public key whose fingerprint is CDAE 0424 F653 F23F 79D1 3ECC EB4D 2AF1 7C32 FE6B

To claim this, I am signing this object:

from __future__ import division
from math import sin, sqrt
from matplotlib import pyplot as plt
n = 5 # n = the energy of the particle.
L = 100 # L = the length of the "box"
PI = 3.14 # Sweet delicious pi
cords = [(x, sqrt(2/L) * sin((PI*n)/L * x)) for x in range(L + 1)]