Re: Numeric root-finding in Python
On Sun, 12 Feb 2012 10:20:17 -0500, inq1ltd inq1...@inqvista.com wrote: I don't know the first thing about this math problem however, if I were to code this I might try ; except ZeroDivisionError: assert w = -1 rather than; except ZeroDivisionError: assert w == -1 Why? DM -- http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On 21/02/2012 01:33, David Monaghan wrote: On Sun, 12 Feb 2012 10:20:17 -0500, inq1ltdinq1...@inqvista.com wrote: I don't know the first thing about this math problem however, if I were to code this I might try ; except ZeroDivisionError: assert w = -1 rather than; except ZeroDivisionError: assert w == -1 Why? DM Why not is probably a better question, given that Dennis Lee Bieber and Dave Angel have already pointed out that this is not legal Python, it'll give a syntax error. -- Cheers. Mark Lawrence. -- http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On Feb 12, 7:41 am, Steven D'Aprano steve
+comp.lang.pyt...@pearwood.info wrote:
This is only peripherally a Python problem, but in case anyone has any
good ideas I'm going to ask it.
I have a routine to calculate an approximation of Lambert's W function,
and then apply a root-finding technique to improve the approximation.
This mostly works well, but sometimes the root-finder gets stuck in a
cycle.
Here's my function:
import math
def improve(x, w, exp=math.exp):
Use Halley's method to improve an estimate of W(x) given
an initial estimate w.
try:
for i in range(36): # Max number of iterations.
ew = exp(w)
a = w*ew - x
b = ew*(w + 1)
err = -a/b # Estimate of the error in the current w.
if abs(err) = 1e-16:
break
print '%d: w= %r err= %r' % (i, w, err)
# Make a better estimate.
c = (w + 2)*a/(2*w + 2)
delta = a/(b - c)
w -= delta
else:
raise RuntimeError('calculation failed to converge', err)
except ZeroDivisionError:
assert w == -1
return w
Here's an example where improve() converges very quickly:
py improve(-0.36, -1.222769842388856)
0: w= -1.222769842388856 err= -2.9158979924038895e-07
1: w= -1.2227701339785069 err= 8.4638038491998997e-16
-1.222770133978506
That's what I expect: convergence in only a few iterations.
Here's an example where it gets stuck in a cycle, bouncing back and forth
between two values:
py improve(-0.36787344117144249, -1.0057222396915309)
0: w= -1.0057222396915309 err= 2.6521238905750239e-14
1: w= -1.0057222396915044 err= -2.6521238905872001e-14
2: w= -1.0057222396915309 err= 2.6521238905750239e-14
3: w= -1.0057222396915044 err= -2.6521238905872001e-14
4: w= -1.0057222396915309 err= 2.6521238905750239e-14
5: w= -1.0057222396915044 err= -2.6521238905872001e-14
[...]
32: w= -1.0057222396915309 err= 2.6521238905750239e-14
33: w= -1.0057222396915044 err= -2.6521238905872001e-14
34: w= -1.0057222396915309 err= 2.6521238905750239e-14
35: w= -1.0057222396915044 err= -2.6521238905872001e-14
Traceback (most recent call last):
File stdin, line 1, in module
File stdin, line 19, in improve
RuntimeError: ('calculation failed to converge', -2.6521238905872001e-14)
(The correct value for w is approximately -1.00572223991.)
I know that Newton's method is subject to cycles, but I haven't found any
discussion about Halley's method and cycles, nor do I know what the best
approach for breaking them would be. None of the papers on calculating
the Lambert W function that I have found mentions this.
Does anyone have any advice for solving this?
--
Steven
Looks like floating point issues to me, rather than something
intrinsic to the iterative algorithm. Surely there is not complex
chaotic behavior to be found in this fairly smooth function in a +/-
1e-14 window. Otoh, there is a lot of floating point significant bit
loss issues to be suspected in the kind of operations you are
performing (exp(x) + something, always a tricky one).
I would start by asking: How accurate is good enough? If its not good
enough, play around the the ordering of your operations, try solving a
transformed problem less sensitive to loss of significance; and begin
by trying different numeric types to see if the problem is sensitive
thereto to begin with.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
在 2012年2月12日星期日UTC+8下午2时41分20秒，Steven D#39;Aprano写道：
This is only peripherally a Python problem, but in case anyone has any
good ideas I'm going to ask it.
I have a routine to calculate an approximation of Lambert's W function,
and then apply a root-finding technique to improve the approximation.
This mostly works well, but sometimes the root-finder gets stuck in a
cycle.
Here's my function:
import math
def improve(x, w, exp=math.exp):
Use Halley's method to improve an estimate of W(x) given
an initial estimate w.
try:
for i in range(36): # Max number of iterations.
ew = exp(w)
a = w*ew - x
W*EXP(W) can converge for negative values of W
b = ew*(w + 1)
b=exp(W)*W+W
err = -a/b # Estimate of the error in the current w.
What's X not expalained?
if abs(err) = 1e-16:
break
print '%d: w= %r err= %r' % (i, w, err)
# Make a better estimate.
c = (w + 2)*a/(2*w + 2)
delta = a/(b - c)
w -= delta
else:
raise RuntimeError('calculation failed to converge', err)
except ZeroDivisionError:
assert w == -1
return w
Here's an example where improve() converges very quickly:
py improve(-0.36, -1.222769842388856)
0: w= -1.222769842388856 err= -2.9158979924038895e-07
1: w= -1.2227701339785069 err= 8.4638038491998997e-16
-1.222770133978506
That's what I expect: convergence in only a few iterations.
Here's an example where it gets stuck in a cycle, bouncing back and forth
between two values:
py improve(-0.36787344117144249, -1.0057222396915309)
0: w= -1.0057222396915309 err= 2.6521238905750239e-14
1: w= -1.0057222396915044 err= -2.6521238905872001e-14
2: w= -1.0057222396915309 err= 2.6521238905750239e-14
3: w= -1.0057222396915044 err= -2.6521238905872001e-14
4: w= -1.0057222396915309 err= 2.6521238905750239e-14
5: w= -1.0057222396915044 err= -2.6521238905872001e-14
[...]
32: w= -1.0057222396915309 err= 2.6521238905750239e-14
33: w= -1.0057222396915044 err= -2.6521238905872001e-14
34: w= -1.0057222396915309 err= 2.6521238905750239e-14
35: w= -1.0057222396915044 err= -2.6521238905872001e-14
Traceback (most recent call last):
File stdin, line 1, in module
File stdin, line 19, in improve
RuntimeError: ('calculation failed to converge', -2.6521238905872001e-14)
(The correct value for w is approximately -1.00572223991.)
I know that Newton's method is subject to cycles, but I haven't found any
discussion about Halley's method and cycles, nor do I know what the best
approach for breaking them would be. None of the papers on calculating
the Lambert W function that I have found mentions this.
Does anyone have any advice for solving this?
--
Steven
I sugest you can use Taylor's series expansion to speed up w*exp(w)
for negative values of w.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On 12/02/2012 10:10, Eelco wrote:
On Feb 12, 7:41 am, Steven D'Apranosteve
+comp.lang.pyt...@pearwood.info wrote:
This is only peripherally a Python problem, but in case anyone has any
good ideas I'm going to ask it.
I have a routine to calculate an approximation of Lambert's W function,
and then apply a root-finding technique to improve the approximation.
This mostly works well, but sometimes the root-finder gets stuck in a
cycle.
Here's my function:
import math
def improve(x, w, exp=math.exp):
Use Halley's method to improve an estimate of W(x) given
an initial estimate w.
try:
for i in range(36): # Max number of iterations.
ew = exp(w)
a = w*ew - x
b = ew*(w + 1)
err = -a/b # Estimate of the error in the current w.
if abs(err)= 1e-16:
break
print '%d: w= %r err= %r' % (i, w, err)
# Make a better estimate.
c = (w + 2)*a/(2*w + 2)
delta = a/(b - c)
w -= delta
else:
raise RuntimeError('calculation failed to converge', err)
except ZeroDivisionError:
assert w == -1
return w
Here's an example where improve() converges very quickly:
py improve(-0.36, -1.222769842388856)
0: w= -1.222769842388856 err= -2.9158979924038895e-07
1: w= -1.2227701339785069 err= 8.4638038491998997e-16
-1.222770133978506
That's what I expect: convergence in only a few iterations.
Here's an example where it gets stuck in a cycle, bouncing back and forth
between two values:
py improve(-0.36787344117144249, -1.0057222396915309)
0: w= -1.0057222396915309 err= 2.6521238905750239e-14
1: w= -1.0057222396915044 err= -2.6521238905872001e-14
2: w= -1.0057222396915309 err= 2.6521238905750239e-14
3: w= -1.0057222396915044 err= -2.6521238905872001e-14
4: w= -1.0057222396915309 err= 2.6521238905750239e-14
5: w= -1.0057222396915044 err= -2.6521238905872001e-14
[...]
32: w= -1.0057222396915309 err= 2.6521238905750239e-14
33: w= -1.0057222396915044 err= -2.6521238905872001e-14
34: w= -1.0057222396915309 err= 2.6521238905750239e-14
35: w= -1.0057222396915044 err= -2.6521238905872001e-14
Traceback (most recent call last):
File stdin, line 1, inmodule
File stdin, line 19, in improve
RuntimeError: ('calculation failed to converge', -2.6521238905872001e-14)
(The correct value for w is approximately -1.00572223991.)
I know that Newton's method is subject to cycles, but I haven't found any
discussion about Halley's method and cycles, nor do I know what the best
approach for breaking them would be. None of the papers on calculating
the Lambert W function that I have found mentions this.
Does anyone have any advice for solving this?
--
Steven
Looks like floating point issues to me, rather than something
intrinsic to the iterative algorithm. Surely there is not complex
chaotic behavior to be found in this fairly smooth function in a +/-
1e-14 window. Otoh, there is a lot of floating point significant bit
loss issues to be suspected in the kind of operations you are
performing (exp(x) + something, always a tricky one).
I would start by asking: How accurate is good enough? If its not good
enough, play around the the ordering of your operations, try solving a
transformed problem less sensitive to loss of significance; and begin
by trying different numeric types to see if the problem is sensitive
thereto to begin with.
HTH.
c:\Users\Mark\Pythontype sda.py
import decimal
def improve(x, w, exp=decimal.Decimal.exp):
Use Halley's method to improve an estimate of W(x) given
an initial estimate w.
try:
for i in range(36): # Max number of iterations.
ew = exp(w)
a = w*ew - x
b = ew*(w + 1)
err = -a/b # Estimate of the error in the current w.
if abs(err) = 1e-16:
break
print '%d: w= %r err= %r' % (i, w, err)
# Make a better estimate.
c = (w + 2)*a/(2*w + 2)
delta = a/(b - c)
w -= delta
else:
raise RuntimeError('calculation failed to converge', err)
print '%d: w= %r err= %r' % (i, w, err)
except ZeroDivisionError:
assert w == -1
return w
improve(decimal.Decimal('-0.36'), decimal.Decimal('-1.222769842388856'))
improve(decimal.Decimal('-0.36787344117144249'),
decimal.Decimal('-1.0057222396915309'))
c:\Users\Mark\Pythonsda.py
0: w= Decimal('-1.222769842388856') err=
Decimal('-2.915897982757542086414504607E-7')
1: w= Decimal('-1.222770133978505953034526059') err=
Decimal('-1.084120148360381932277303211E-19')
0: w= Decimal('-1.0057222396915309') err=
Decimal('5.744538819905061986438230561E-15')
1: w= Decimal('-1.005722239691525155461180092') err= Decimal('-0E+2')
--
Cheers.
Mark Lawrence.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On 2/12/12 6:41 AM, Steven D'Aprano wrote: This is only peripherally a Python problem, but in case anyone has any good ideas I'm going to ask it. I have a routine to calculate an approximation of Lambert's W function, and then apply a root-finding technique to improve the approximation. This mostly works well, but sometimes the root-finder gets stuck in a cycle. I don't have any advice for fixing your code, per se, but I would just grab mpmath and use their lambertw function: http://mpmath.googlecode.com/svn/trunk/doc/build/functions/powers.html#lambert-w-function -- Robert Kern I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth. -- Umberto Eco -- http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
I don't know the first thing about this math problem however,
if I were to code this I might try ;
except ZeroDivisionError:
assert w = -1
rather than;
except ZeroDivisionError:
assert w == -1
jimonlinux
On Sunday, February 12, 2012 06:41:20 AM Steven D'Aprano wrote:
This is only peripherally a Python problem, but in case anyone has any
good ideas I'm going to ask it.
I have a routine to calculate an approximation of Lambert's W function,
and then apply a root-finding technique to improve the approximation.
This mostly works well, but sometimes the root-finder gets stuck in a
cycle.
Here's my function:
import math
def improve(x, w, exp=math.exp):
Use Halley's method to improve an estimate of W(x) given
an initial estimate w.
try:
for i in range(36): # Max number of iterations.
ew = exp(w)
a = w*ew - x
b = ew*(w + 1)
err = -a/b # Estimate of the error in the current w.
if abs(err) = 1e-16:
break
print '%d: w= %r err= %r' % (i, w, err)
# Make a better estimate.
c = (w + 2)*a/(2*w + 2)
delta = a/(b - c)
w -= delta
else:
raise RuntimeError('calculation failed to converge', err)
except ZeroDivisionError:
assert w == -1
return w
Here's an example where improve() converges very quickly:
py improve(-0.36, -1.222769842388856)
0: w= -1.222769842388856 err= -2.9158979924038895e-07
1: w= -1.2227701339785069 err= 8.4638038491998997e-16
-1.222770133978506
That's what I expect: convergence in only a few iterations.
Here's an example where it gets stuck in a cycle, bouncing back and forth
between two values:
py improve(-0.36787344117144249, -1.0057222396915309)
0: w= -1.0057222396915309 err= 2.6521238905750239e-14
1: w= -1.0057222396915044 err= -2.6521238905872001e-14
2: w= -1.0057222396915309 err= 2.6521238905750239e-14
3: w= -1.0057222396915044 err= -2.6521238905872001e-14
4: w= -1.0057222396915309 err= 2.6521238905750239e-14
5: w= -1.0057222396915044 err= -2.6521238905872001e-14
[...]
32: w= -1.0057222396915309 err= 2.6521238905750239e-14
33: w= -1.0057222396915044 err= -2.6521238905872001e-14
34: w= -1.0057222396915309 err= 2.6521238905750239e-14
35: w= -1.0057222396915044 err= -2.6521238905872001e-14
Traceback (most recent call last):
File stdin, line 1, in module
File stdin, line 19, in improve
RuntimeError: ('calculation failed to converge', -2.6521238905872001e-14)
(The correct value for w is approximately -1.00572223991.)
I know that Newton's method is subject to cycles, but I haven't found any
discussion about Halley's method and cycles, nor do I know what the best
approach for breaking them would be. None of the papers on calculating
the Lambert W function that I have found mentions this.
Does anyone have any advice for solving this?
--
http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On 02/12/2012 10:20 AM, inq1ltd wrote: I don't know the first thing about this math problem however, if I were to code this I might try ; except ZeroDivisionError: assert w = -1 You top-posted. Please type your response after whatever you're quoting. In my case, I only need a portion of what you said, and my remarks are following it. assert takes an expression, so the one above is just wrong. Fortunately, Python would tell you with SyntaxError: invalid syntax -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On Feb 12, 6:41 am, Steven D'Aprano steve +comp.lang.pyt...@pearwood.info wrote: err = -a/b # Estimate of the error in the current w. if abs(err) = 1e-16: break If the result you're expecting is around -1.005, this exit condition is rather optimistic: the difference between the two Python floats either side of this value is already 2.22e-16, so you're asking for less than half a ulp of error! As to the rest; your error estimate simply doesn't have enough precision. The main problem is in the computation of a, where you're subtracting two almost identical values. The absolute error incurred in computing w*exp(w) is of the same order of magnitude as the difference 'w*exp(w) - x' itself, so err has lost essentially all of its significant bits, and is at best only a crude indicator of the size of the error. The solution would be to compute the quantities 'exp(w), w*exp(w), and w*exp(w) - x' all with extended precision. For the other quantities, there shouldn't be any major issues---after all, you only need a few significant bits of 'delta' for it to be useful, but with the subtraction that generates a, you don't even get those few significant bits. (The correct value for w is approximately -1.00572223991.) Are you sure? Wolfram Alpha gives me the following value for W(-1, -0.36787344117144249455719773322925902903079986572265625): -1.005722239691522978... so it looks as though the values you're getting are at least alternating around the exact value. -- Mark -- http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On 2/12/2012 5:10 AM, Eelco wrote:
On Feb 12, 7:41 am, Steven D'Apranosteve
+comp.lang.pyt...@pearwood.info wrote:
This is only peripherally a Python problem, but in case anyone has any
good ideas I'm going to ask it.
I have a routine to calculate an approximation of Lambert's W function,
and then apply a root-finding technique to improve the approximation.
This mostly works well, but sometimes the root-finder gets stuck in a
cycle.
Here's my function:
import math
def improve(x, w, exp=math.exp):
Use Halley's method to improve an estimate of W(x) given
an initial estimate w.
try:
for i in range(36): # Max number of iterations.
ew = exp(w)
a = w*ew - x
b = ew*(w + 1)
err = -a/b # Estimate of the error in the current w.
if abs(err)= 1e-16:
break
print '%d: w= %r err= %r' % (i, w, err)
# Make a better estimate.
c = (w + 2)*a/(2*w + 2)
delta = a/(b - c)
w -= delta
else:
raise RuntimeError('calculation failed to converge', err)
except ZeroDivisionError:
assert w == -1
return w
Here's an example where improve() converges very quickly:
py improve(-0.36, -1.222769842388856)
0: w= -1.222769842388856 err= -2.9158979924038895e-07
1: w= -1.2227701339785069 err= 8.4638038491998997e-16
-1.222770133978506
That's what I expect: convergence in only a few iterations.
Here's an example where it gets stuck in a cycle, bouncing back and forth
between two values:
py improve(-0.36787344117144249, -1.0057222396915309)
0: w= -1.0057222396915309 err= 2.6521238905750239e-14
1: w= -1.0057222396915044 err= -2.6521238905872001e-14
2: w= -1.0057222396915309 err= 2.6521238905750239e-14
3: w= -1.0057222396915044 err= -2.6521238905872001e-14
4: w= -1.0057222396915309 err= 2.6521238905750239e-14
35: w= -1.0057222396915044 err= -2.6521238905872001e-14
Traceback (most recent call last):
File stdin, line 1, inmodule
File stdin, line 19, in improve
RuntimeError: ('calculation failed to converge', -2.6521238905872001e-14)
(The correct value for w is approximately -1.00572223991.)
I know that Newton's method is subject to cycles, but I haven't found any
discussion about Halley's method and cycles, nor do I know what the best
approach for breaking them would be. None of the papers on calculating
the Lambert W function that I have found mentions this.
Does anyone have any advice for solving this?
Looks like floating point issues to me, rather than something
intrinsic to the iterative algorithm. Surely there is not complex
chaotic behavior to be found in this fairly smooth function in a +/-
1e-14 window. Otoh, there is a lot of floating point significant bit
loss issues to be suspected in the kind of operations you are
performing (exp(x) + something, always a tricky one).
To investigate this, I would limit the iterations to 2 or 3 and print
ew, a,b,c, and delta, maybe in binary(hex) form
I would start by asking: How accurate is good enough? If its not good
enough, play around the the ordering of your operations, try solving a
transformed problem less sensitive to loss of significance; and begin
by trying different numeric types to see if the problem is sensitive
thereto to begin with.
--
Terry Jan Reedy
--
http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On Sun, 12 Feb 2012 13:52:48 +, Robert Kern wrote: I don't have any advice for fixing your code, per se, but I would just grab mpmath and use their lambertw function: That's no fun! I'd never see mpmath before, it looks like it is worth investigating. Nevertheless, I still intend working on my lambert function, as it's a good learning exercise. I did look into SciPy's lambert too, and was put off by this warning in the docs: In some corner cases, lambertw might currently fail to converge http://docs.scipy.org/doc/scipy/reference/generated/ scipy.special.lambertw.html Naturally I thought I can do better than that. Looks like I can't :) -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On Sun, 12 Feb 2012 12:18:15 -0800, Mark Dickinson wrote: On Feb 12, 6:41 am, Steven D'Aprano steve +comp.lang.pyt...@pearwood.info wrote: err = -a/b # Estimate of the error in the current w. if abs(err) = 1e-16: break If the result you're expecting is around -1.005, this exit condition is rather optimistic: the difference between the two Python floats either side of this value is already 2.22e-16, so you're asking for less than half a ulp of error! I was gradually coming to the conclusion on my own that I was being overly optimistic with my error condition, although I couldn't put it into words *why*. Thanks for this Mark, this is exactly the sort of thing I need to learn -- as is obvious, I'm no expert on numeric programming. As to the rest; your error estimate simply doesn't have enough precision. The main problem is in the computation of a, where you're subtracting two almost identical values. The absolute error incurred in computing w*exp(w) is of the same order of magnitude as the difference 'w*exp(w) - x' itself, so err has lost essentially all of its significant bits, and is at best only a crude indicator of the size of the error. The solution would be to compute the quantities 'exp(w), w*exp(w), and w*exp(w) - x' all with extended precision. Other than using Decimal, there's no way to do that in pure Python, is there? We have floats (double) and that's it. For the other quantities, there shouldn't be any major issues---after all, you only need a few significant bits of 'delta' for it to be useful, but with the subtraction that generates a, you don't even get those few significant bits. (The correct value for w is approximately -1.00572223991.) Are you sure? Wolfram Alpha gives me the following value for W(-1, -0.36787344117144249455719773322925902903079986572265625): -1.005722239691522978... I did say *approximately*. The figure I quote comes from my HP-48GX, and seems to be accurate to the precision offered by the HP. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: Numeric root-finding in Python
On 02/12/2012 06:05 PM, Steven D'Aprano wrote: On Sun, 12 Feb 2012 12:18:15 -0800, Mark Dickinson wrote: On Feb 12, 6:41 am, Steven D'Apranosteve +comp.lang.pyt...@pearwood.info wrote: err = -a/b # Estimate of the error in the current w. if abs(err)= 1e-16: break If the result you're expecting is around -1.005, this exit condition is rather optimistic: the difference between the two Python floats either side of this value is already 2.22e-16, so you're asking for less than half a ulp of error! I was gradually coming to the conclusion on my own that I was being overly optimistic with my error condition, although I couldn't put it into words *why*. Thanks for this Mark, this is exactly the sort of thing I need to learn -- as is obvious, I'm no expert on numeric programming. me either. But comments below. As to the rest; your error estimate simply doesn't have enough precision. The main problem is in the computation of a, where you're subtracting two almost identical values.SNIP SNIP Two pieces of my history that come to mind. 40+ years ago I got a letter from a user of our computer stating that our math seemed to be imprecise in certain places. He was very polte about it, and admitted to maybe needing a different algorithm. The letter was so polite that I (as author of the math microcode) worked on his problem, and found the difficulty, as well as a solution. The problem was figuring out the difference in a machining table between being level every place on its surface (in which case it would be slightly curved to match the earth), or being perfectly flat (in which case some parts of the table would be further from the earth's center than others) The table was 200 feet long, and we were talking millionths of an inch. He solved it three ways, and got three different answers. The first two differed in the 3rd place, which he thought far too big an error, and the third answer was just about exactly half the others. Well the 2:1 discrepancy just happens when you change your assumption of what part of the flat table is level. If the center is level, then the edges are only 100 feet out, while if the edge is level, the other edge is 200 feet out. But the other solution was very interesting. Turns out he sketched a right triangle, with narrow angle at the center of the earth, side opposite being 200 feet. He then calculated the difference between the other two sides. one 8000 miles, and the other 8000 miles plus a few microinches. He got that distance by subtracting the sine from the tangent, or something similar to that. I had microcoded both those functions, and was proud of their accuracy. But if you subtract two 13 digit numbers that only differ in the last 3, you only get 3 digits worth of accuracy, best case. Solution was to apply some similar triangles, and some trivial approximations, and the problem turned out not to need trig at all, and accurate to at least 12 places. if I recall, it was something like 8000mi is to 200 feet, as 200 feet is to X. Cross multiply and it's just arithmetic. The other problem was even earlier. It was high school physics, and the challenge was to experimentally determine the index of refraction of air to 5 places. Problem is our measurements can't be that accurate. So this is the same thing in reverse. Find a way to measure the difference of the index of refraction of air and vacuum, to one or two places, and add that to 1. taken together with lots of other experience, i try to avoid commiting an algorithm to code before thinking about errors, convergence, and exceptional conditions. I've no experience with Lambert, but I suspect it can be attacked similarly. -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
Numeric root-finding in Python
This is only peripherally a Python problem, but in case anyone has any
good ideas I'm going to ask it.
I have a routine to calculate an approximation of Lambert's W function,
and then apply a root-finding technique to improve the approximation.
This mostly works well, but sometimes the root-finder gets stuck in a
cycle.
Here's my function:
import math
def improve(x, w, exp=math.exp):
Use Halley's method to improve an estimate of W(x) given
an initial estimate w.
try:
for i in range(36): # Max number of iterations.
ew = exp(w)
a = w*ew - x
b = ew*(w + 1)
err = -a/b # Estimate of the error in the current w.
if abs(err) = 1e-16:
break
print '%d: w= %r err= %r' % (i, w, err)
# Make a better estimate.
c = (w + 2)*a/(2*w + 2)
delta = a/(b - c)
w -= delta
else:
raise RuntimeError('calculation failed to converge', err)
except ZeroDivisionError:
assert w == -1
return w
Here's an example where improve() converges very quickly:
py improve(-0.36, -1.222769842388856)
0: w= -1.222769842388856 err= -2.9158979924038895e-07
1: w= -1.2227701339785069 err= 8.4638038491998997e-16
-1.222770133978506
That's what I expect: convergence in only a few iterations.
Here's an example where it gets stuck in a cycle, bouncing back and forth
between two values:
py improve(-0.36787344117144249, -1.0057222396915309)
0: w= -1.0057222396915309 err= 2.6521238905750239e-14
1: w= -1.0057222396915044 err= -2.6521238905872001e-14
2: w= -1.0057222396915309 err= 2.6521238905750239e-14
3: w= -1.0057222396915044 err= -2.6521238905872001e-14
4: w= -1.0057222396915309 err= 2.6521238905750239e-14
5: w= -1.0057222396915044 err= -2.6521238905872001e-14
[...]
32: w= -1.0057222396915309 err= 2.6521238905750239e-14
33: w= -1.0057222396915044 err= -2.6521238905872001e-14
34: w= -1.0057222396915309 err= 2.6521238905750239e-14
35: w= -1.0057222396915044 err= -2.6521238905872001e-14
Traceback (most recent call last):
File stdin, line 1, in module
File stdin, line 19, in improve
RuntimeError: ('calculation failed to converge', -2.6521238905872001e-14)
(The correct value for w is approximately -1.00572223991.)
I know that Newton's method is subject to cycles, but I haven't found any
discussion about Halley's method and cycles, nor do I know what the best
approach for breaking them would be. None of the papers on calculating
the Lambert W function that I have found mentions this.
Does anyone have any advice for solving this?
--
Steven
--
http://mail.python.org/mailman/listinfo/python-list
