code /home/ryan/Dropbox/Studies/QuantProject/Current/Python-Quant/ & disownWriting and presenting.
The Matrix Exponential is implemented in areas of:
- Graph Centrality modelling cite:parkPowerWalkRevisiting2013
- Systems of Linear Differential Equations Ch. 8.4
- Theory of Algebraic Lie Groups Ch. 2
However the method to implement matrix exponentiation provided by the documentation cite:MatricesSymPyDocumentation2020 and referenced in the development repository cite:MatrixExponentialIssue2019 does not appear to be implemented very well, for example the following provides a very long result:
from __future__ import division
from sympy import *
x, y, z, t = symbols('x y z t')
k, m, n = symbols('k m n', integer=True)
f, g, h = symbols('f g h', cls=Function)
init_printing()
init_printing(use_latex='mathjax', latex_mode='equation')
import pyperclip
def lx(expr):
pyperclip.copy(latex(expr))
print(expr)A = Matrix([
[11, 12, 13],
[21, 22, 23],
[31, 32, 33]
])
expr = exp(A)
expr.doit()$$ { \scriptsize \left[\begin{matrix}- \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{26}{- \sqrt{1149} - 22}\right) - \frac{1}{\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}} \left(-1 - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{26}{- \sqrt{1149} - 22} - \frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 2\right)\right) \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)} - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\right) + \frac{1}{\left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right) e-33 + \sqrt{1149}} \left(-1 - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{26}{- \sqrt{1149} - 22} - \frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 2\right)\right) \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right) + 1 + \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(-1 - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{26}{- \sqrt{1149} - 22} - \frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 2\right)\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right) + \frac{2}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}}\right) e33 + \sqrt{1149} & - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) + \frac{1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)} - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\right) \left(\frac{13}{- \sqrt{1149} - 22} - \frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 1\right) - \frac{1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right) e-33 + \sqrt{1149}} \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right) \left(\frac{13}{- \sqrt{1149} - 22} - \frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 1\right) + \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(\frac{1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right)2 \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right) \left(\frac{13}{- \sqrt{1149} - 22} - \frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 1\right) + \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}}\right) e33 + \sqrt{1149} & \frac{1}{\left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right) e^{-33 + \sqrt{1149}}} \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right) - \frac{1}{\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}} \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)} - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\right) - \frac{e33 + \sqrt{1149}}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\-2 - \frac{1}{\left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right) e-33 + \sqrt{1149}} \left(-1 - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{26}{- \sqrt{1149} - 22} - \frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 2\right)\right) \left(- \frac{273}{-22 + \sqrt{1149}} + 23\right) + \frac{2}{\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}} \left(-1 - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{26}{- \sqrt{1149} - 22} - \frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 2\right)\right) \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)} - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\right) + \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{13104}{- \sqrt{1149} - 22} + 1104}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{52}{- \sqrt{1149} - 22}\right) - \frac{e^{33 + \sqrt{1149}}}{- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}} \left(- \frac{1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(-1 - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{26}{- \sqrt{1149} - 22} - \frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 2\right)\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right) + \frac{2}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}}\right) \left(- \frac{273}{- \sqrt{1149} - 22} + 23\right) & \frac{\left(- \frac{273}{-22 + \sqrt{1149}} + 23\right) \left(\frac{13}{- \sqrt{1149} - 22} - \frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 1\right)}{\left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right) \left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right) e^{-33 + \sqrt{1149}}} - \frac{2}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)} - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\right) \left(\frac{13}{- \sqrt{1149} - 22} - \frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 1\right) + \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{26}{- \sqrt{1149} - 22}\right) - \frac{e33 + \sqrt{1149}}{- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}} \left(\frac{1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right)2 \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right) \left(\frac{13}{- \sqrt{1149} - 22} - \frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 1\right) + \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}}\right) \left(- \frac{273}{- \sqrt{1149} - 22} + 23\right) & \frac{1}{\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}} \left(- \frac{26}{-22 + \sqrt{1149}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)} - \frac{2}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\right) - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{\left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right) e^{-33 + \sqrt{1149}}} + \frac{e33 + \sqrt{1149}}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(- \frac{273}{- \sqrt{1149} - 22} + 23\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\- \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{26}{- \sqrt{1149} - 22}\right) - \frac{1}{\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}} \left(-1 - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{26}{- \sqrt{1149} - 22} - \frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 2\right)\right) \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)} - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\right) + \frac{1}{\left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right) e-33 + \sqrt{1149}} \left(-1 - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{26}{- \sqrt{1149} - 22} - \frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 2\right)\right) + 1 + \left(- \frac{1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(-1 - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{26}{- \sqrt{1149} - 22} - \frac{- \frac{6552}{- \sqrt{1149} - 22} + 552}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 2\right)\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right) + \frac{2}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}}\right) e33 + \sqrt{1149} & - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) + \frac{1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)} - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\right) \left(\frac{13}{- \sqrt{1149} - 22} - \frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 1\right) - \frac{\frac{13}{- \sqrt{1149} - 22} - \frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right) e-33 + \sqrt{1149}} + \left(\frac{1}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right)2 \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right) \left(\frac{13}{- \sqrt{1149} - 22} - \frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} + 1\right) + \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}}\right) e33 + \sqrt{1149} & \frac{1}{\left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right) e^{-33 + \sqrt{1149}}} - \frac{1}{\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}} \left(- \frac{13}{-22 + \sqrt{1149}} + \frac{- \frac{3276}{-22 + \sqrt{1149}} + 276}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)} - \frac{1}{- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}} \left(\frac{- \frac{3276}{- \sqrt{1149} - 22} + 276}{\left(- \sqrt{1149} - 22\right) \left(- \frac{59 \sqrt{1149}}{95} - \frac{1837}{95}\right)} - \frac{13}{- \sqrt{1149} - 22}\right) \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\right) - \frac{e33 + \sqrt{1149}}{\left(- \frac{33}{94} + \frac{5 \sqrt{1149}}{94}\right) \left(\frac{33 \sqrt{1149}}{2303} + \frac{5745}{2303}\right)} \left(- \frac{26}{-22 + \sqrt{1149}} - \frac{- \frac{273}{-22 + \sqrt{1149}} + 23}{- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}} + \frac{- \frac{6552}{-22 + \sqrt{1149}} + 552}{\left(-22 + \sqrt{1149}\right) \left(- \frac{1837}{95} + \frac{59 \sqrt{1149}}{95}\right)}\right)\end{matrix}\right] } $$ Simplifying this result doesn’t seem to help either:
simplify(expr)Methods suggested online only provide numerical solutions or partial sums:
- python - Sympy Symbolic Matrix Exponential - Stack Overflow
- python - Exponentiate symbolic matrix expression using SymPy - Stack Overflow
- Calculate state transition matrix in python - Stack Overflow
Instead this will need to be implemented from first principles.
A Matrix Exponential is defined by using the ordinary exponential power series Ch. 2,Ch. 8.4 (should we prove the power series generally?):
\begin{align} e\mathit{\mathbf{X}} = ∑∞k= 0 \left[ \frac{1}{k!} ⋅ \mathit{\mathbf{X^k}} \right] \end{align}
This definition can be expanded upon however by using properties of logarithms:
\begin{align}
b &= elog_e{\left( b \right) }, \quad ∀ b ∈ \mathbb{C} \label{eq:bydef}
\implies b\mathbf{X}&= \left( elog_e{\left( b \right) } \right)\mathbf{X} \label{eq:tojustify} \
\implies b\mathbf{X} &= elog_e{b \mathbf{X} }
\end{align}
The identity in \eqref{eq:bydef} is justified by the definition of the complex log. However some discussion is required for \eqref{eq:tojustify} because it is not clear that the exponential will generally distribute throught he parenthesis like so $\left( a⋅ b \right)k = a^k⋅ b^k$, for example consider $\left( \left[ - 1 \right]^2 ⋅ 3 \right)\frac{1{2}} ≠ \left[ - 1 \right]\frac{2{2}} ⋅ 3\frac{1{2}}$.
A sufficient condition for this identity is $k ∈ \mathbb{Z}*$, consider this example which will be important later:
\begin{align} \left( log_e{\left( b \right) }\mathbf{X} \right)k , \quad ∀ k ∈ \mathbb{Z*} \end{align}
Because multiplication is commutative
\begin{align}
\left( log_e{\left( b \right) }\mathbf{X} \right)k &= \underbrace{log_e{\left( b \right) }⋅ log_e{\left( b \right) } ⋅ log_e{\left( b \right) }\ldots }k \text{ times} × \underbrace{\mathbf{X}\mathbf{X}\mathbf{X}\ldots}k \text{ times} \notag
&= log_e^k{\left( b \right) } \mathbf{X}^k \label{eq:matpower}
\end{align}
Now consider the the following by applying \eqref{eq:matpower}:
\begin{align}
eX&= ∑∞k= 0 \left[ \frac{1}{k!} \mathbf{X}k \right] \notag
\implies ebX&= ∑∞k= 0 \left[ \frac{1}{k!} \left( b\mathbf{X} \right)k \right] \quad ∀ b ∈ \mathbb{C} \notag \
&= ∑∞k= 0 \left[ \frac{1}{k!}b^k \mathbf{X}^k \right] \notag \
&= \left( e^b \right)\mathbf{X} \notag \
&\implies eb \mathbf{X} = e\mathbf{Xb}= \left( e^b \right)\mathbf{X}= \left( e\mathbf{X} \right)^b \qquad \qquad \square \label{eq:expmatpower}
\end{align}
So the matrix exponential for an arbitrary base could be given by:
\begin{align}
b = elog_e{\left( b \right) }, \quad ∀ b ∈ \mathbb{C} \notag
\implies b\mathbf{X} &= \left( elog_e{\left( b \right) } \right)\mathbf{X} \notag \
\text{as per \eqref{eq:expmatpower}} \notag \
b\mathbf{X} &= elog_e{\left( b \right) {\mathbf{X}}} \notag \
b\mathbf{X} &= ∑∞k= 0 \left[ \frac{\left( log_e{\left( b \right) }\mathbf{X} \right)^k}{k!} \right] \notag \
&= ∑∞k= 0 \left[ \frac{log_e k{\left( b \right) }}{k!}\mathbf{X}k \right]
\end{align}
This is also consistent with the McLaurin Series expansion of $b\mathbf{X} \enspace (∀ b ∈ \mathbb{C})$:
\begin{align*}
f\left( x \right) &= ∑∞k= 0 \left[ \frac{f\left( n \right)\left( 0 \right)}{k!} xk \right]
\implies b^x &= ∑∞k= 0 \left[ \frac{\frac{\mathrm{d}^n }{\mathrm{d} x^n}\left( b^x \right) |x=0 }{k!} x^k \right] \
\implies b\mathbf{X} &= ∑∞k= 0 \left[ \frac{\frac{\mathrm{d}^n }{\mathrm{d}\mathbf{X}^n } \left( b\mathbf{X} \right) |\mathbf{X= \mathbf{O}}}{k!} \mathbf{X}^k \right]
\end{align*}
By ordinary calculus identities we have$f\left( x \right) = bx \implies f\left( n \right)\left( x \right) = bx log_e^n{\left( b \right)}$ which distribute through a matrix and hence:
\begin{align*}
b^x &= ∑∞k= 0 \left[ \frac{b^0 log_e^k{\left( b \right) }}{k!} x^k \right]
\implies b\mathbf{X} &= ∑∞k= 0 \left[ \frac{b^0 log_e^k{\left( b \right) }}{k!} \mathbf{X}^k \right]
\end{align*}
By the previous identity:
\begin{align*}
\implies b\mathbf{X} &= ∑∞k= 0 {\left[ \frac{{\left( log_e{\left( b \right) } \mathbf{X} \right)}^k}{k!} \right]}
&= elog_e{\left( b \right) \mathbf{X}}
\end{align*}
Matrix-Matrix exponentiation has applications in quantum mechanics p. 84.
As for Matrices with the requirements:
- Square
- Normal:
- Commutes with it’s congugate transpose
- Non Singular
- Non Zero Determinant
\begin{align*}
\left| \left| A-I \right| \right|<1 &\implies elog_e{\left( \mathbf{A \right) }} = \mathbf{A} \enspace \text{(By Lie Groups Springer Textbook)}
&\implies \mathbf{A}\mathbf{B} =\left( elog_e{\left( \mathbf{A \right) }} \right)\mathbf{B} \
& \text{Similar justification as \eqref{eq:expmatpower}} \
& \implies \mathbf{A}\mathbf{B}= elog_e{\left( \mathbf{A \right) } \mathbf{B}}
\end{align*}
However the following identities are by Definition anyway: cite:barradasIteratedExponentiationMatrixMatrix1994
\begin{align}
\mathbf{A}\mathbf{B}&= elog_e{\left( \mathbf{A \right) } \mathbf{B}}
\ \mathbf{B} \mathbf{A} &= e \mathbf{B log_e{\left( \mathbf{A} \right) } }
\end{align}
def matexp(mat, base = E):
"""
Return the Matrix Exponential of a square matrix
"""
import copy
import sympy
# Should realy test for sympy vs numpy array
# Test for Square Matrix
if mat.shape[0] != mat.shape[1]:
print("ERROR: Only defined for Square matrices")
return
m = zeros(mat.shape[0])
for i in range(m.shape[0]):
for j in range(m.shape[1]):
m[i,j] = Sum((mat[i,j]*ln(base))**k/factorial(k), (k, 0, oo)).doit()
return mmatexp(A, pi)But it would be nice to expand this to matrix bases for there uses in quantum mechanics.
The built in method for a**mat is not implemented.
there is exp(mat) but this returns garbage (see github issue), (see other solution on stack exchange that is numeric and example)
show our method with proofs of
cauchy power taylor then exp
then show our code
A = Matrix([ [11,12,13], [21,22,23], [31,32,33] ])
B = Matrix([
[1,2,3],
[4,5,6],
[7,8,9]
])
A**B[fn:1]
Park, Laurence A. F., and Simeon Simoff. “Power Walk: Revisiting the Random Surfer.” In Proceedings of the 18th Australasian Document Computing Symposium, 50–57. ADCS ‘13. Brisbane, Queensland, Australia: Association for Computing Machinery, 2013. https://doi.org/10.1145/2537734.2537749.
[fn:2] Zill, Dennis G., and Michael R. Cullen. “8.4 Matrix Exponential.” In Differential Equations with Boundary-Value Problems, 7th ed. Belmont, CA: Brooks/Cole, Cengage Learning, 2009.
[fn:3] Hall, Brian C. Lie Groups, Lie Algebras, and Representations: An Elementary Introduction. Second edition. Graduate Texts in Mathematics 222. Cham ; New York: Springer, 2015.
[fn:4] Hall, Brian C. Lie Groups, Lie Algebras, and Representations: An Elementary Introduction. Second edition. Graduate Texts in Mathematics 222. Cham ; New York: Springer, 2015.
[fn:5] Zill, Dennis G., and Michael R. Cullen. “8.4 Matrix Exponential.” In Differential Equations with Boundary-Value Problems, 7th ed. Belmont, CA: Brooks/Cole, Cengage Learning, 2009.
[fn:6] Hall, Brian C. Lie Groups, Lie Algebras, and Representations: An Elementary Introduction. Second edition. Graduate Texts in Mathematics 222. Cham ; New York: Springer, 2015.
[fn:7] Barradas, I., and J. E. Cohen. “Iterated Exponentiation, Matrix-Matrix Exponentiation, and Entropy.” Journal of Mathematical Analysis and Applications 183, no. 1 (April 1, 1994): 76–88. https://doi.org/10.1006/jmaa.1994.1132.
[fn:8] Barradas, I., and J. E. Cohen. “Iterated Exponentiation, Matrix-Matrix Exponentiation, and Entropy.” Journal of Mathematical Analysis and Applications 183, no. 1 (April 1, 1994): 76–88. https://doi.org/10.1006/jmaa.1994.1132.
A linear recurrence relation is of the form:
\begin{align} ∑∞n= 0 \left[ c_i ⋅ a_n \right] = 0, \quad ∃ c ∈ \mathbb{R}, \enspace ∀ i<k∈\mathbb{Z}^+ \label{eq:recurrence-relation-def} \end{align}
In order to find a solution for
\begin{align} g: \left( a_n \right)n∈\mathbb{Z^+} → \mathbb{C}\left[ \left[ x \right] \right]: g\left( a_n \right) = ∑∞n= 0\left[ \frac{x^n}{n!} a_n \right] \label{eq:gen-func-def} \end{align}
This technique is referred to as generating functions. cite:lehmanReadingsMathematicsComputer2010
A Generating Function, as shown in eqref:eq:gen-func-def is a way of encoding an infinite series of numbers (
\begin{align*}
a_0= 1
a_0= 1 \
an+ 2 = an+ 1 + 2 a_n, \quad n \geq 0
\end{align*}
To solve this we can use what’s known as a Generating Function, see the disucssion below
We will make consider the function
\begin{align} f\left( x \right)= ∑∞n= 0 \left[ a_nx^n \right] \label{eq:pow-gen-func-np0} \end{align}
It can be shown (see eqref:iterate-pow-gen-func) that:
\begin{align}
∑∞n= 0 \left[ an+ 1 x^n \right] &= \frac{f\left( x \right)- a_0}{x} \label{eq:pow-gen-func-np1}
∑∞n= 0 \left[ an+ 2 x^n \right] &= \frac{f\left( x \right) - a_0 - a_1x}{x^2} \label{eq:pow-gen-func-np2}
\end{align}
So to use the generating Function consider:
\begin{align}
2a_n + an+ 1 &= an+ 2 \nonumber
2a_nx^n + an+ 1 x^n &= an+ 2 x^n \nonumber \
∑∞n= 0 \left[ 2a_nx^n \right] + ∑∞n= 0 \left[ an+ 1 x^n \right] &= ∑∞n= 0 \left[ an+ 2 x^n \right] \label{eq:series-rep-pow-example}
\end{align}
Observe that in eqref:eq:series-rep-pow-exampluse tuse te
By applying the previous identity shown in eqref:eq:pow-gen-func-np0, eqref:eq:pow-gen-func-np1 and eqref:eq:pow-gen-func-np2:
\begin{align}
2f\left( x \right) + \frac{f\left( x \right)- a_0}{x} &= \frac{f\left( x \right)- a_0}{- a_1x}x^2 \nonumber
\implies f\left( x \right) &= \frac{1}{1- x- x^2} \label{eq:power-series-form-example}
\end{align}
| WARNING |
|---|
| I accidently dropped the |
The function
By partial fractions it is known:
\begin{align*}
f\left( x \right)&= \frac{1}{1- x- x^2}
&= \frac{- 1}{x^2 + x - 1}\
&= \frac{- 1}{\left( x- 2 \right)\left( x- 1 \right)}\
&= \frac{A_1}{x- 2}+ \frac{A_2}{x- 1}, \quad A_i ∈ \mathbb{R}, i ∈ \mathbb{Z}^+ \
\implies - 1 &= A_1\left( x- 1 \right) + A_2\left( x- 2 \right)\
\text{Let
- 1&= A_1\left( 2-1 \right) + 0
&= A_1 = - 1
\text{Let
- 1 &= 0 + A_2 \left( 1- 2 \right)
\implies A_2&= 1
\text{Hence:}\
f\left( x \right)&= \frac{1}{x- 1} - \frac{1}{x- 2}
\end{align*}
By definition of the power series:
\begin{align} ∑∞n= 0\left[ rx^n \right] = \frac{1}{1- rx^n} \label{eq:pow-series-definition} \end{align}
we can conclude that:
\begin{align*}
\frac{1}{x- 1}&= -\frac{1}{1 -\left( 1 \right) x}
&= -∑∞n= 0\left[ x^n \right] \
\frac{-1}{x- 2} &= \frac{1}{2- x} \
&= \frac{1}{2}\frac{1}{1-\frac{1}{2}x} \
&= \frac{1}{2} ∑∞n= 0\left[ \left( \frac{1}{2}x \right) ^n \right]
\end{align*}
and so:
\begin{align*}
f\left( x \right) &= \frac{1}{2}∑∞n= 0\left[ \left( \frac{1}{2}x \right) ^n \right] - ∑∞n= 0\left[ x^n \right]
f\left( x \right) &= ∑∞n= 0\left[ \frac{1}{2}\left( \frac{1}{2}x \right) ^n -x^n \right] \
f\left( x \right) &= ∑∞n= 0\left[ \frac{1}{2 ⋅ 2^n} x^n -x^n \right] \
f\left( x \right) &= ∑∞n= 0\left[x^n {\left( {\frac{1}{2 ⋅ 2^n} -1} \right) } \right] \
\end{align*}
\begin{align}
\implies a_n &= \frac{1}{2 ⋅ 2^n} - 1 \label{eq:seq-end-value}
\end{align}
Consider the Fibonacci Sequence:
\begin{align}
an&= an - 1 + an - 2 \nonumber
\iff an+ 2 &= an+ 1 + a_n \label{eq:fib-def}
\end{align}
Solving this outright is quite difficult, a power series generating function can be used to solve it as shown in section #power-series-gen-func-example, which provides a corresponding equation to the effect of:
\begin{align*} x^2 f\left( x \right) - x f \left( x \right) - f\left( x \right)= 0 \end{align*}
This however still requires a little intuition, however, just from observation, this appears similar in structure to the following ordinary differential equation, which would be fairly easy to deal with:
\begin{align*} f”\left( x \right)- f’\left( x \right)- f\left( x \right)= 0 \end{align*}
This would imply that $f\left( x \right) \propto emx, \quad ∃ m ∈ \mathbb{Z}$ because
\begin{align*} f\left( x \right)= erx = ∑∞n= 0 \left[ r \frac{x^n}{n!} \right] \end{align*}
So this would give an easy means by which to solve the linear recurrence relation.
Consider using the following generating function, (the derivative of the generating function as in eqref:eq:exp-gen-def-2 and eqref:eq:exp-gen-def-3 is provided in section #Derivative-exp-gen-function)
\begin{alignat}{2}
f \left( x \right) &= ∑∞n= 0 \left[ an ⋅ \frac{x^n}{n!} \right] &= e^x \label{eq:exp-gen-def-1}
f’\left( x \right) &= ∑∞n= 0 \left[ an+1 ⋅ \frac{x^n}{n!} \right] &= e^x \label{eq:exp-gen-def-2} \
f”\left( x \right) &= ∑∞n= 0 \left[ an+2 ⋅ \frac{x^n}{n!} \right] &= e^x \label{eq:exp-gen-def-3}
\end{alignat}
So the recursive relation from ref:eq:fib-def could be expressed :
\begin{align*}
an+ 2 &= an+ 1 + an
\frac{x^n}{n!} an+ 2 &= \frac{x^n}{n!}\left( an+ 1 + an \right)\
∑∞n= 0 \left[ \frac{x^n}{n!} an+ 2 \right] &= ∑∞n= 0 \left[ \frac{x^n}{n!} an+ 1 \right] + ∑∞n= 0 \left[ \frac{x^n}{n!} an \right] \
f”\left( x \right) &= f’\left( x \right)+ f\left( x \right)
\end{align*}
Using the theory of higher order linear differential equations with constant coefficients it can be shown:
\begin{align*} f\left( x \right)= c_1 ⋅ \mathrm{exp}\left[ \left( \frac{1- \sqrt{5} }{2} \right)x \right] + c_2 ⋅ \mathrm{exp}\left[ \left( \frac{1 + \sqrt{5} }{2} \right) \right] \end{align*}
By equating this to the power series:
\begin{align*} f\left( x \right)&= ∑∞n= 0 \left[ \left( c_1\left( \frac{1- \sqrt{5} }{2} \right)^n + c_2 ⋅ \left( \frac{1+ \sqrt{5} }{2} \right)^n \right) ⋅ \frac{x^n}{n} \right] \end{align*}
Now given that:
\begin{align*} f\left( x \right)= ∑∞n= 0 \left[ a_n \frac{x^n}{n!} \right] \end{align*}
We can conclude that:
\begin{align*} a_n = c_1⋅ \left( \frac{1- \sqrt{5} }{2} \right)^n + c_2 ⋅ \left( \frac{1+ \sqrt{5} }{2} \right) \end{align*}
By applying the initial conditions:
\begin{align*}
a_0= c_1 + c_2 \implies c_1= - c_2
a_1= c_1 \left( \frac{1+ \sqrt{5} }{2} \right) - c_1 \frac{1-\sqrt{5} }{2} \implies c_1 = \frac{1}{\sqrt{5} }
\end{align*}
And so finally we have the solution to the Fibonacci Sequence ref:eq:fib-def:
\begin{align}
a_n &= \frac{1}{\sqrt{5} } \left[ \left( \frac{1+ \sqrt{5} }{2} \right)^n - \left( \frac{1- \sqrt{5} }{2} \right)^n \right] \nonumber
&= \frac{\varphi^n - ψ^n}{\sqrt{5} } \nonumber\
&=\frac{\varphi^n - ψ^n}{\varphi - ψ} \label{eq:fib-sol}
\end{align}
where:
$\varphi = \frac{1+ \sqrt{5} }{2} ≈ 1.61\ldots$ $ψ = 1-\varphi = \frac{1- \sqrt{5} }{2} ≈ 0.61\ldots$
\begin{align}
f\left( x \right) &= ∑∞n= 0 \left[ a_n \frac{x^n}{n!} \right] \label{eq:exp-pow-series}
f’\left( x \right)) &= \frac{\mathrm{d} }{\mathrm{d} x}\left( ∑∞n= 0 \left[ a_n \frac{x^n}{n!} \right] \right) \nonumber \
&= \frac{\mathrm{d}}{\mathrm{d} x} \left( a_0 \frac{x^0}{0!} + a_1 \frac{x^1}{1!} + a_2 \frac{x^2}{2!}+ a_3 \frac{x^3}{3! } + \ldots \frac{x^k}{k!} \right) \nonumber \
&= ∑∞n= 0 \left[ \frac{\mathrm{d} }{\mathrm{d} x}\left( a_n \frac{x^n}{n!} \right) \right] \nonumber \
&= ∑∞n= 0 {\left[{ \frac{a_n}{{\left({ n- 1 }\right)!}} } xn- 1 \right]} \nonumber \
\implies f’(x) &= ∑∞n= 0 {\left[{ \frac{x^n}{n!}an+ 1 }\right]} \label{eq:exp-pow-series-sol}
\end{align}
If $f\left( x \right)= ∑∞n= 0 \left[ a_n \frac{x^n}{n!} \right]$ can it be shown by induction that $\frac{\mathrm{d}^k }{\mathrm{d} x^k} \left( f\left( x \right) \right)= fk \left( x \right) ∑∞n= 0 \left[ x^n \frac{an+ k}{n!} \right]$
An equation of the form:
\begin{align} ∑∞n=0 \left[ ci ⋅ f(n)(x) \right] = 0 \label{eq:hom-ode} \end{align}
is said to be a homogenous linear ODE:
- Linear
- because the equation is linear with respect to $f(x)$
- Ordinary
- because there are no partial derivatives (e.g. $\frac{∂ }{∂ x}{\left({ f{\left({ x }\right)} }\right)}$ )
- Differential
- because the derivates of the function are concerned
- Homogenous
- because the /RHS/ is 0
- A non-homogeous equation would have a non-zero RHS
There will be
- $f(x)=ci ⋅ em_{ix}$
- $f(x)=ci ⋅ xj⋅ em_{ix}$
where:
- $∑ki=0\left[ cimk-i \right] = 0$
- This is referred to the characteristic equation of the recurrence relation or ODE cite:levinSolvingRecurrenceRelations2018
- $∃ i,j ∈ \mathbb{Z}+ ∩ \left[0,k\right]$
- These is often referred to as repeated roots cite:levinSolvingRecurrenceRelations2018,zillMatrixExponential2009 with a multiplicity corresponding to the number of repetitions of that root \textsection 3.2
An example of a recurrence relation with all unique roots is the fibonacci sequence, as described in section #solving-the-sequence.
Consider the linear recurrence relation eqref:eq:recurrence-relation-def:
\begin{align} ∑∞n= 0 \left[ c_i ⋅ a_n \right] = 0, \quad ∃ c ∈ \mathbb{R}, \enspace ∀ i<k∈\mathbb{Z}^+ \tag{$\textrm{\ref{eq:recurrence-relation-def}}^2$} \end{align}
By implementing the exponential generating function as shown in eqref:eq:exp-gen-def-1, this provides:
\begin{align}
∑ki= 0 {\left[{ c_i ⋅ a_n } \right]} = 0 \nonumber
\intertext{By Multiplying through and summing: } \notag \
\implies ∑ki= 0 {\left[{ ∑∞n= 0 {\left[{ c_i a_n \frac{x^n}{n!} }\right]} }\right]} \nonumber = 0 \
∑ki= 0 {\left[{ c_i ∑∞n= 0 {\left[{ a_n \frac{x^n}{n!} }\right]} }\right]} \nonumber = 0 \
\end{align}
Recall from eqref:eq:exp-gen-def-1 the generating function
\begin{align} ∑ki= 0 {\left[{ c_i f{\left({ k \right)} } } {\left({ x }\right)} \right]} \label{eq:exp-gen-def-proof} &= 0 \end{align}
Now assume that the solution exists and all roots of the characteristic polynomial are unique (i.e. the solution is of the form $f{\left({ x }\right)} \propto em_i x: \quad m_i ≠ m_j ∀ i≠ j$), this implies that Ch. 4 :
\begin{align} f{\left({ x }\right)} = ∑ki= 0 {\left[{ k_i em_i x }\right]}, \quad ∃ m,k ∈ \mathbb{C} \nonumber \end{align}
This can be re-expressed in terms of the exponential power series, in order to relate the solution of the function
\begin{align}
∑ki= 0 {\left[{ k_i em_i x }\right]} &= ∑ki= 0 {\left[{ k_i ∑∞n= 0 \frac{{\left({ m_i x }\right)}^n}{n!} }\right]} \nonumber
&= ∑ki= 0 ∑∞n= 0 k_i m_i^n \frac{x^n}{n!} \nonumber\
&= ∑∞n= 0 ∑ki= 0 k_i m_i^n \frac{x^n}{n!} \nonumber \
&= ∑∞n= 0 {\left[{ \frac{x^n}{n!} ∑ki=0 {\left[{ k_im^n_i }\right]} }\right]}, \quad ∃ k_i ∈ \mathbb{C}, \enspace ∀ i ∈ \mathbb{Z}^+∩ {\left[{ 1, k }\right]} \label{eq:unique-root-sol-power-series-form}
\end{align}
Recall the definition of the generating function from ref:eq:exp-gen-def-proof, by relating this to eqref:eq:unique-root-sol-power-series-form:
\begin{align}
f{\left({ x }\right)} &= ∑∞n= 0 {\left[{ \frac{x^n}{n!} a_n }\right]} \nonumber
&= ∑∞n= 0 {\left[{ \frac{x^n}{n!} ∑ki=0 {\left[{ k_im^n_i }\right]} }\right]} \nonumber \
\implies a_n &= ∑kn= 0 {\left[{ k_im_i^n }\right]} \nonumber \ \nonumber
\square
\end{align}
This can be verified by the fibonacci sequence as shown in section #solving-the-sequence, the solution to the characteristic equation is
\begin{alignat}{4}
f{\left({ x }\right)} &= &c_1 e\varphi x + &c_2 e{\left({ 1-\varphi \right)} x}, \quad &∃ c_1, c_2 ∈ \mathbb{R} ⊂ \mathbb{C} \nonumber
\iff a_n &= &k_1 n\varphi + &k_2 n1- \varphi, &∃ k_1, k_2 ∈ \mathbb{R} ⊂ \mathbb{C} \nonumber
\end{alignat}
Consider the following recurrence relation:
\begin{align}
a_n - 10an+ 1 + 25an+ 2&= 0 \label{eq:hom-repeated-roots-recurrence}
\implies ∑∞n= 0 {\left[{ a_n \frac{x^n}{n!} }\right]} - 10 ∑∞n= 0 {\left[{ \frac{x^n}{n!}+ }\right]} + 25 ∑∞n= 0 {\left[{ an+ 2 \frac{x^n}{n!} }\right]}&= 0 \nonumber
\end{align}
By applying the definition of the exponential generating function at eqref:eq:exp-gen-def-1 :
\begin{align} f”{\left({ x }\right)}- 10f’{\left({ x }\right)}+ 25f{\left({ x }\right)}= 0 \nonumber \label{eq:rep-roots-func-ode} \end{align}
By implementing the already well-established theory of linear ODE’s, the characteristic equation for eqref:eq:rep-roots-func-ode can be expressed as:
\begin{align}
m^2- 10m+ 25 = 0 \nonumber
{\left({ m- 5 }\right)}^2 = 0 \nonumber \
m= 5 \label{eq:rep-roots-recurrence-char-sol}
\end{align}
Herein lies a complexity, in order to solve this, the solution produced from eqref:eq:rep-roots-recurrence-char-sol can be used with the Reduction of Order technique to produce a solution that will be of the form \textsection 4.3.
\begin{align} f{\left({ x }\right)}= c_1e5x + c_2 x e5x \label{eq:rep-roots-ode-sol} \end{align}
eqref:eq:rep-roots-ode-sol can be expressed in terms of the exponential power series in order to try and relate the solution for the function back to the generating function, observe however the following power series identity (TODO Prove this in section #prove-ext-exp-power-series-rep-roots):
\begin{align} x^ke^x &= ∑∞n= 0 {\left[{ \frac{x^n}{{\left({ n- k }\right)}!} }\right]}, \quad ∃ k ∈ \mathbb{Z}^+ \label{eq:uniq-roots-pow-series-ident} \end{align}
by applying identity eqref:eq:uniq-roots-pow-series-ident to equation eqref:eq:rep-roots-ode-sol
\begin{align}
\implies f{\left({ x }\right)} &= ∑∞n= 0 {\left[{ c_1 \frac{{\left({ 5x }\right)}^n}{n!} }\right]} + ∑∞n= 0 {\left[{ c_2 n \frac{{\left({ 5x^n }\right)}}{n{\left({ n-1 }\right)}!} }\right]} \nonumber
&= ∑∞n= 0 {\left[{ \frac{x^n}{n!} {\left({ c15^n + c_2 n 5^n }\right)} }\right]} \nonumber
\end{align}
Given the defenition of the exponential generating function from eqref:eq:exp-gen-def-1
\begin{align}
f{\left({ x }\right)}&= ∑∞n= 0 {\left[{ a_n \frac{x^n}{n!} }\right]} \nonumber
\iff a_n &= c15^n + c_2n_5^n \nonumber \ \nonumber
\ \nonumber \
\square \nonumber
\end{align}
In order to prove the the solution for a $k\mathrm{th}$ order recurrence relation with
Consider a recurrence relation of the form:
\begin{align}
∑kn= 0 {\left[{ c_i a_n }\right]} = 0 \nonumber
\implies ∑∞n= 0 ∑ki= 0 c_i a_n \frac{x^n}{n!} = 0 \nonumber \
∑ki= 0 ∑∞n= 0 c_i a_n \frac{x^n}{n!} \nonumber
\end{align}
By substituting for the value of the generating function (from eqref:eq:exp-gen-def-1):
\begin{align} ∑ki= 0 {\left[{ c_if{\left({ k \right)}} {\left({ x }\right)} }\right]} \label{eq:gen-form-rep-roots-ode} \end{align}
Assume that eqref:eq:gen-form-rep-roots-ode corresponds to a charecteristic polynomial with only 1 root of multiplicity
\begin{align}
& ∑ki= 0 {\left[{ c_i m^i }\right]} = 0 ∧ m=B, \enspace ∃! B ∈ \mathbb{C} \nonumber
\implies f{\left({ x }\right)}&= ∑ki= 0 {\left[{ x^i A_i emx }\right]}, \quad ∃ A ∈ \mathbb{C}^+, \enspace ∀ i ∈ {\left[{ 1,k }\right]} ∩ \mathbb{N} \label{eq:sol-rep-roots-ode} \
\end{align}
Recall the following power series identity (proved in section xxx):
\begin{align} x^k e^x = ∑∞n= 0 {\left[{ \frac{x^n}{{\left({ n- k }\right)}!} }\right]} \nonumber \end{align}
By applying this to eqref:eq:sol-rep-roots-ode :
\begin{align}
f{\left({ x }\right)}&= ∑ki= 0 {\left[{ A_i ∑∞n= 0 {\left[{ \frac{{\left({ x m }\right)}^n}{{\left({ n- i }\right)}!} }\right]} }\right]} \nonumber
&= ∑∞n= 0 {\left[{ ∑ki=0 {\left[{ \frac{x^n}{n!} \frac{n!}{{\left({ n- i }\right)}} A_i m^n }\right]} }\right]} # \
&= ∑∞n= 0 {\left[{ \frac{x^n}{n!} ∑ki=0 {\left[{ \frac{n!}{{\left({ n- i }\right)}} A_i m^n }\right]} }\right]}
\end{align}
Recall the generating function that was used to get ref:eq:gen-form-rep-roots-ode:
\begin{align}
f{\left({ x }\right)}&= ∑∞n= 0 {\left[{ a_n \frac{x^n}{n!} }\right]} \nonumber
\implies a_n &= ∑ki= 0 {\left[{ A_i \frac{n!}{{\left({ n- i }\right)}!} m^n }\right]} \nonumber \
&= ∑ki= 0 {\left[{ m^n A_i ∏0k {\left[{ n- {\left({ i- 1 }\right)} }\right]} }\right]}
& \intertext{$\because \enspace i \leq k$} \notag \
&= ∑ki= 0 {\left[{ A_i^* m^n n^i }\right]}, \quad ∃ A_i ∈ \mathbb{C}, \enspace ∀ i\leqk ∈ \mathbb{Z}^+ \nonumber \
\ \nonumber \
\square \nonumber
\end{align}
In sections #uniq-roots-recurrence and *Unique Roots of Characteristic Equation it was shown that a recurrence relation can be related to an ODE and then that solution can be transformed to provide a solution for the recurrence relation, when the charecteristic polynomial has either complex roots or 1 repeated root. Generally the solution to a linear ODE will be a superposition of solutions for each root, repeated or unique and so here it will be shown that these two can be combined and that the solution will still hold.
Consider a Recursive relation with constant coefficients:
This can be expressed in terms of the exponential generating function:
- Use the Generating function to get an ODE
- The ODE will have a solution that is a combination of the above two forms
- The solution will translate back to a combination of both above forms
- https://math.stackexchange.com/a/1775226
- https://math.stackexchange.com/a/593553
- https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/IvanNiven.pdf
Misc
- https://brilliant.org/wiki/generating-functions-solving-recurrence-relations/
- https://www.math.cmu.edu/~af1p/Teaching/Combinatorics/Slides/Generating-Functions.pdf
- https://www.math.cmu.edu/~af1p/Teaching/Combinatorics/Slides/Generating-Functions.pdf
When