532 lines
18 KiB
Python
532 lines
18 KiB
Python
from ..libmp.backend import xrange
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# TODO: should use diagonalization-based algorithms
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class MatrixCalculusMethods(object):
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def _exp_pade(ctx, a):
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"""
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Exponential of a matrix using Pade approximants.
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See G. H. Golub, C. F. van Loan 'Matrix Computations',
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third Ed., page 572
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TODO:
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- find a good estimate for q
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- reduce the number of matrix multiplications to improve
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performance
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"""
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def eps_pade(p):
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return ctx.mpf(2)**(3-2*p) * \
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ctx.factorial(p)**2/(ctx.factorial(2*p)**2 * (2*p + 1))
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q = 4
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extraq = 8
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while 1:
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if eps_pade(q) < ctx.eps:
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break
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q += 1
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q += extraq
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j = int(max(1, ctx.mag(ctx.mnorm(a,'inf'))))
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extra = q
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prec = ctx.prec
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ctx.dps += extra + 3
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try:
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a = a/2**j
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na = a.rows
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den = ctx.eye(na)
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num = ctx.eye(na)
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x = ctx.eye(na)
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c = ctx.mpf(1)
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for k in range(1, q+1):
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c *= ctx.mpf(q - k + 1)/((2*q - k + 1) * k)
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x = a*x
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cx = c*x
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num += cx
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den += (-1)**k * cx
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f = ctx.lu_solve_mat(den, num)
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for k in range(j):
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f = f*f
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finally:
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ctx.prec = prec
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return f*1
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def expm(ctx, A, method='taylor'):
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r"""
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Computes the matrix exponential of a square matrix `A`, which is defined
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by the power series
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.. math ::
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\exp(A) = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \ldots
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With method='taylor', the matrix exponential is computed
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using the Taylor series. With method='pade', Pade approximants
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are used instead.
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**Examples**
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Basic examples::
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>>> from mpmath import *
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>>> mp.dps = 15; mp.pretty = True
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>>> expm(zeros(3))
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[1.0 0.0 0.0]
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[0.0 1.0 0.0]
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[0.0 0.0 1.0]
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>>> expm(eye(3))
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[2.71828182845905 0.0 0.0]
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[ 0.0 2.71828182845905 0.0]
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[ 0.0 0.0 2.71828182845905]
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>>> expm([[1,1,0],[1,0,1],[0,1,0]])
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[ 3.86814500615414 2.26812870852145 0.841130841230196]
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[ 2.26812870852145 2.44114713886289 1.42699786729125]
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[0.841130841230196 1.42699786729125 1.6000162976327]
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>>> expm([[1,1,0],[1,0,1],[0,1,0]], method='pade')
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[ 3.86814500615414 2.26812870852145 0.841130841230196]
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[ 2.26812870852145 2.44114713886289 1.42699786729125]
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[0.841130841230196 1.42699786729125 1.6000162976327]
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>>> expm([[1+j, 0], [1+j,1]])
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[(1.46869393991589 + 2.28735528717884j) 0.0]
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[ (1.03776739863568 + 3.536943175722j) (2.71828182845905 + 0.0j)]
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Matrices with large entries are allowed::
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>>> expm(matrix([[1,2],[2,3]])**25)
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[5.65024064048415e+2050488462815550 9.14228140091932e+2050488462815550]
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[9.14228140091932e+2050488462815550 1.47925220414035e+2050488462815551]
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The identity `\exp(A+B) = \exp(A) \exp(B)` does not hold for
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noncommuting matrices::
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>>> A = hilbert(3)
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>>> B = A + eye(3)
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>>> chop(mnorm(A*B - B*A))
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0.0
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>>> chop(mnorm(expm(A+B) - expm(A)*expm(B)))
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0.0
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>>> B = A + ones(3)
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>>> mnorm(A*B - B*A)
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1.8
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>>> mnorm(expm(A+B) - expm(A)*expm(B))
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42.0927851137247
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"""
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if method == 'pade':
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prec = ctx.prec
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try:
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A = ctx.matrix(A)
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ctx.prec += 2*A.rows
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res = ctx._exp_pade(A)
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finally:
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ctx.prec = prec
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return res
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A = ctx.matrix(A)
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prec = ctx.prec
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j = int(max(1, ctx.mag(ctx.mnorm(A,'inf'))))
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j += int(0.5*prec**0.5)
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try:
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ctx.prec += 10 + 2*j
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tol = +ctx.eps
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A = A/2**j
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T = A
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Y = A**0 + A
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k = 2
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while 1:
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T *= A * (1/ctx.mpf(k))
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if ctx.mnorm(T, 'inf') < tol:
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break
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Y += T
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k += 1
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for k in xrange(j):
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Y = Y*Y
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finally:
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ctx.prec = prec
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Y *= 1
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return Y
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def cosm(ctx, A):
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r"""
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Gives the cosine of a square matrix `A`, defined in analogy
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with the matrix exponential.
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Examples::
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>>> from mpmath import *
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>>> mp.dps = 15; mp.pretty = True
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>>> X = eye(3)
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>>> cosm(X)
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[0.54030230586814 0.0 0.0]
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[ 0.0 0.54030230586814 0.0]
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[ 0.0 0.0 0.54030230586814]
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>>> X = hilbert(3)
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>>> cosm(X)
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[ 0.424403834569555 -0.316643413047167 -0.221474945949293]
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[-0.316643413047167 0.820646708837824 -0.127183694770039]
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[-0.221474945949293 -0.127183694770039 0.909236687217541]
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>>> X = matrix([[1+j,-2],[0,-j]])
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>>> cosm(X)
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[(0.833730025131149 - 0.988897705762865j) (1.07485840848393 - 0.17192140544213j)]
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[ 0.0 (1.54308063481524 + 0.0j)]
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"""
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B = 0.5 * (ctx.expm(A*ctx.j) + ctx.expm(A*(-ctx.j)))
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if not sum(A.apply(ctx.im).apply(abs)):
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B = B.apply(ctx.re)
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return B
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def sinm(ctx, A):
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r"""
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Gives the sine of a square matrix `A`, defined in analogy
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with the matrix exponential.
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Examples::
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>>> from mpmath import *
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>>> mp.dps = 15; mp.pretty = True
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>>> X = eye(3)
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>>> sinm(X)
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[0.841470984807897 0.0 0.0]
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[ 0.0 0.841470984807897 0.0]
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[ 0.0 0.0 0.841470984807897]
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>>> X = hilbert(3)
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>>> sinm(X)
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[0.711608512150994 0.339783913247439 0.220742837314741]
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[0.339783913247439 0.244113865695532 0.187231271174372]
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[0.220742837314741 0.187231271174372 0.155816730769635]
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>>> X = matrix([[1+j,-2],[0,-j]])
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>>> sinm(X)
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[(1.29845758141598 + 0.634963914784736j) (-1.96751511930922 + 0.314700021761367j)]
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[ 0.0 (0.0 - 1.1752011936438j)]
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"""
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B = (-0.5j) * (ctx.expm(A*ctx.j) - ctx.expm(A*(-ctx.j)))
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if not sum(A.apply(ctx.im).apply(abs)):
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B = B.apply(ctx.re)
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return B
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def _sqrtm_rot(ctx, A, _may_rotate):
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# If the iteration fails to converge, cheat by performing
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# a rotation by a complex number
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u = ctx.j**0.3
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return ctx.sqrtm(u*A, _may_rotate) / ctx.sqrt(u)
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def sqrtm(ctx, A, _may_rotate=2):
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r"""
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Computes a square root of the square matrix `A`, i.e. returns
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a matrix `B = A^{1/2}` such that `B^2 = A`. The square root
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of a matrix, if it exists, is not unique.
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**Examples**
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Square roots of some simple matrices::
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>>> from mpmath import *
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>>> mp.dps = 15; mp.pretty = True
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>>> sqrtm([[1,0], [0,1]])
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[1.0 0.0]
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[0.0 1.0]
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>>> sqrtm([[0,0], [0,0]])
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[0.0 0.0]
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[0.0 0.0]
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>>> sqrtm([[2,0],[0,1]])
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[1.4142135623731 0.0]
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[ 0.0 1.0]
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>>> sqrtm([[1,1],[1,0]])
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[ (0.920442065259926 - 0.21728689675164j) (0.568864481005783 + 0.351577584254143j)]
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[(0.568864481005783 + 0.351577584254143j) (0.351577584254143 - 0.568864481005783j)]
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>>> sqrtm([[1,0],[0,1]])
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[1.0 0.0]
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[0.0 1.0]
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>>> sqrtm([[-1,0],[0,1]])
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[(0.0 - 1.0j) 0.0]
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[ 0.0 (1.0 + 0.0j)]
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>>> sqrtm([[j,0],[0,j]])
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[(0.707106781186547 + 0.707106781186547j) 0.0]
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[ 0.0 (0.707106781186547 + 0.707106781186547j)]
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A square root of a rotation matrix, giving the corresponding
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half-angle rotation matrix::
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>>> t1 = 0.75
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>>> t2 = t1 * 0.5
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>>> A1 = matrix([[cos(t1), -sin(t1)], [sin(t1), cos(t1)]])
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>>> A2 = matrix([[cos(t2), -sin(t2)], [sin(t2), cos(t2)]])
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>>> sqrtm(A1)
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[0.930507621912314 -0.366272529086048]
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[0.366272529086048 0.930507621912314]
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>>> A2
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[0.930507621912314 -0.366272529086048]
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[0.366272529086048 0.930507621912314]
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The identity `(A^2)^{1/2} = A` does not necessarily hold::
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>>> A = matrix([[4,1,4],[7,8,9],[10,2,11]])
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>>> sqrtm(A**2)
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[ 4.0 1.0 4.0]
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[ 7.0 8.0 9.0]
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[10.0 2.0 11.0]
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>>> sqrtm(A)**2
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[ 4.0 1.0 4.0]
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[ 7.0 8.0 9.0]
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[10.0 2.0 11.0]
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>>> A = matrix([[-4,1,4],[7,-8,9],[10,2,11]])
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>>> sqrtm(A**2)
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[ 7.43715112194995 -0.324127569985474 1.8481718827526]
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[-0.251549715716942 9.32699765900402 2.48221180985147]
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[ 4.11609388833616 0.775751877098258 13.017955697342]
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>>> chop(sqrtm(A)**2)
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[-4.0 1.0 4.0]
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[ 7.0 -8.0 9.0]
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[10.0 2.0 11.0]
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For some matrices, a square root does not exist::
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>>> sqrtm([[0,1], [0,0]])
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Traceback (most recent call last):
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...
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ZeroDivisionError: matrix is numerically singular
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Two examples from the documentation for Matlab's ``sqrtm``::
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>>> mp.dps = 15; mp.pretty = True
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>>> sqrtm([[7,10],[15,22]])
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[1.56669890360128 1.74077655955698]
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[2.61116483933547 4.17786374293675]
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>>>
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>>> X = matrix(\
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... [[5,-4,1,0,0],
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... [-4,6,-4,1,0],
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... [1,-4,6,-4,1],
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... [0,1,-4,6,-4],
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... [0,0,1,-4,5]])
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>>> Y = matrix(\
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... [[2,-1,-0,-0,-0],
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... [-1,2,-1,0,-0],
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... [0,-1,2,-1,0],
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... [-0,0,-1,2,-1],
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... [-0,-0,-0,-1,2]])
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>>> mnorm(sqrtm(X) - Y)
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4.53155328326114e-19
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"""
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A = ctx.matrix(A)
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# Trivial
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if A*0 == A:
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return A
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prec = ctx.prec
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if _may_rotate:
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d = ctx.det(A)
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if abs(ctx.im(d)) < 16*ctx.eps and ctx.re(d) < 0:
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return ctx._sqrtm_rot(A, _may_rotate-1)
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try:
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ctx.prec += 10
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tol = ctx.eps * 128
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Y = A
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Z = I = A**0
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k = 0
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# Denman-Beavers iteration
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while 1:
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Yprev = Y
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try:
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Y, Z = 0.5*(Y+ctx.inverse(Z)), 0.5*(Z+ctx.inverse(Y))
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except ZeroDivisionError:
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if _may_rotate:
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Y = ctx._sqrtm_rot(A, _may_rotate-1)
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break
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else:
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raise
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mag1 = ctx.mnorm(Y-Yprev, 'inf')
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mag2 = ctx.mnorm(Y, 'inf')
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if mag1 <= mag2*tol:
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break
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if _may_rotate and k > 6 and not mag1 < mag2 * 0.001:
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return ctx._sqrtm_rot(A, _may_rotate-1)
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k += 1
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if k > ctx.prec:
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raise ctx.NoConvergence
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finally:
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ctx.prec = prec
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Y *= 1
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return Y
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def logm(ctx, A):
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r"""
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Computes a logarithm of the square matrix `A`, i.e. returns
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a matrix `B = \log(A)` such that `\exp(B) = A`. The logarithm
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of a matrix, if it exists, is not unique.
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**Examples**
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Logarithms of some simple matrices::
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>>> from mpmath import *
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>>> mp.dps = 15; mp.pretty = True
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>>> X = eye(3)
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>>> logm(X)
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[0.0 0.0 0.0]
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[0.0 0.0 0.0]
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[0.0 0.0 0.0]
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>>> logm(2*X)
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[0.693147180559945 0.0 0.0]
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[ 0.0 0.693147180559945 0.0]
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[ 0.0 0.0 0.693147180559945]
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>>> logm(expm(X))
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[1.0 0.0 0.0]
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[0.0 1.0 0.0]
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[0.0 0.0 1.0]
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A logarithm of a complex matrix::
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>>> X = matrix([[2+j, 1, 3], [1-j, 1-2*j, 1], [-4, -5, j]])
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>>> B = logm(X)
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>>> nprint(B)
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[ (0.808757 + 0.107759j) (2.20752 + 0.202762j) (1.07376 - 0.773874j)]
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[ (0.905709 - 0.107795j) (0.0287395 - 0.824993j) (0.111619 + 0.514272j)]
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[(-0.930151 + 0.399512j) (-2.06266 - 0.674397j) (0.791552 + 0.519839j)]
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>>> chop(expm(B))
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[(2.0 + 1.0j) 1.0 3.0]
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[(1.0 - 1.0j) (1.0 - 2.0j) 1.0]
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[ -4.0 -5.0 (0.0 + 1.0j)]
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A matrix `X` close to the identity matrix, for which
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`\log(\exp(X)) = \exp(\log(X)) = X` holds::
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>>> X = eye(3) + hilbert(3)/4
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>>> X
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[ 1.25 0.125 0.0833333333333333]
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[ 0.125 1.08333333333333 0.0625]
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[0.0833333333333333 0.0625 1.05]
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>>> logm(expm(X))
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[ 1.25 0.125 0.0833333333333333]
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[ 0.125 1.08333333333333 0.0625]
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[0.0833333333333333 0.0625 1.05]
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>>> expm(logm(X))
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[ 1.25 0.125 0.0833333333333333]
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[ 0.125 1.08333333333333 0.0625]
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[0.0833333333333333 0.0625 1.05]
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A logarithm of a rotation matrix, giving back the angle of
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the rotation::
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>>> t = 3.7
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>>> A = matrix([[cos(t),sin(t)],[-sin(t),cos(t)]])
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>>> chop(logm(A))
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[ 0.0 -2.58318530717959]
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[2.58318530717959 0.0]
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>>> (2*pi-t)
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2.58318530717959
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For some matrices, a logarithm does not exist::
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>>> logm([[1,0], [0,0]])
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Traceback (most recent call last):
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...
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ZeroDivisionError: matrix is numerically singular
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Logarithm of a matrix with large entries::
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>>> logm(hilbert(3) * 10**20).apply(re)
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[ 45.5597513593433 1.27721006042799 0.317662687717978]
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[ 1.27721006042799 42.5222778973542 2.24003708791604]
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[0.317662687717978 2.24003708791604 42.395212822267]
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"""
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A = ctx.matrix(A)
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prec = ctx.prec
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try:
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ctx.prec += 10
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tol = ctx.eps * 128
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I = A**0
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B = A
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n = 0
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while 1:
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B = ctx.sqrtm(B)
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n += 1
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if ctx.mnorm(B-I, 'inf') < 0.125:
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break
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T = X = B-I
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L = X*0
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k = 1
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while 1:
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if k & 1:
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L += T / k
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else:
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L -= T / k
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T *= X
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if ctx.mnorm(T, 'inf') < tol:
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break
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k += 1
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if k > ctx.prec:
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raise ctx.NoConvergence
|
|
finally:
|
|
ctx.prec = prec
|
|
L *= 2**n
|
|
return L
|
|
|
|
def powm(ctx, A, r):
|
|
r"""
|
|
Computes `A^r = \exp(A \log r)` for a matrix `A` and complex
|
|
number `r`.
|
|
|
|
**Examples**
|
|
|
|
Powers and inverse powers of a matrix::
|
|
|
|
>>> from mpmath import *
|
|
>>> mp.dps = 15; mp.pretty = True
|
|
>>> A = matrix([[4,1,4],[7,8,9],[10,2,11]])
|
|
>>> powm(A, 2)
|
|
[ 63.0 20.0 69.0]
|
|
[174.0 89.0 199.0]
|
|
[164.0 48.0 179.0]
|
|
>>> chop(powm(powm(A, 4), 1/4.))
|
|
[ 4.0 1.0 4.0]
|
|
[ 7.0 8.0 9.0]
|
|
[10.0 2.0 11.0]
|
|
>>> powm(extraprec(20)(powm)(A, -4), -1/4.)
|
|
[ 4.0 1.0 4.0]
|
|
[ 7.0 8.0 9.0]
|
|
[10.0 2.0 11.0]
|
|
>>> chop(powm(powm(A, 1+0.5j), 1/(1+0.5j)))
|
|
[ 4.0 1.0 4.0]
|
|
[ 7.0 8.0 9.0]
|
|
[10.0 2.0 11.0]
|
|
>>> powm(extraprec(5)(powm)(A, -1.5), -1/(1.5))
|
|
[ 4.0 1.0 4.0]
|
|
[ 7.0 8.0 9.0]
|
|
[10.0 2.0 11.0]
|
|
|
|
A Fibonacci-generating matrix::
|
|
|
|
>>> powm([[1,1],[1,0]], 10)
|
|
[89.0 55.0]
|
|
[55.0 34.0]
|
|
>>> fib(10)
|
|
55.0
|
|
>>> powm([[1,1],[1,0]], 6.5)
|
|
[(16.5166626964253 - 0.0121089837381789j) (10.2078589271083 + 0.0195927472575932j)]
|
|
[(10.2078589271083 + 0.0195927472575932j) (6.30880376931698 - 0.0317017309957721j)]
|
|
>>> (phi**6.5 - (1-phi)**6.5)/sqrt(5)
|
|
(10.2078589271083 - 0.0195927472575932j)
|
|
>>> powm([[1,1],[1,0]], 6.2)
|
|
[ (14.3076953002666 - 0.008222855781077j) (8.81733464837593 + 0.0133048601383712j)]
|
|
[(8.81733464837593 + 0.0133048601383712j) (5.49036065189071 - 0.0215277159194482j)]
|
|
>>> (phi**6.2 - (1-phi)**6.2)/sqrt(5)
|
|
(8.81733464837593 - 0.0133048601383712j)
|
|
|
|
"""
|
|
A = ctx.matrix(A)
|
|
r = ctx.convert(r)
|
|
prec = ctx.prec
|
|
try:
|
|
ctx.prec += 10
|
|
if ctx.isint(r):
|
|
v = A ** int(r)
|
|
elif ctx.isint(r*2):
|
|
y = int(r*2)
|
|
v = ctx.sqrtm(A) ** y
|
|
else:
|
|
v = ctx.expm(r*ctx.logm(A))
|
|
finally:
|
|
ctx.prec = prec
|
|
v *= 1
|
|
return v
|