Trouble getting loop to break
Dick Moores
x**0/0! + x**1/1! + x**2/2! + x**3/3! + ... = e**x (x is non-negative)
It works OK for many x, but for many the loop doesn't break. Is there
a way to get it to break where I want it to, i.e., when the sum
equals the limit as closely as the precision allows?
Here's what I have:
======= series_xToN_OverFactorialN.py ==========================
#!/usr/bin/env python
#coding=utf-8
# series_xToN_OverFactorialN.py limit is e**x from p.63 in The
Pleasures of Pi,e
from mpmath import mpf, e, exp, factorial
import math
import time
precision = 100
mpf.dps = precision
n = mpf(0)
x = mpf(raw_input("Enter a non-negative int or float: "))
term = 1
sum = 0
limit = e**x
k = 0
while True:
k += 1
term = x**n/factorial(n)
sum += term
print " sum = %s k = %d" % (sum, k)
print "exp(%s) = %s" % (x, exp(x))
print " e**%s = %s" % (x, e**x)
print
if sum >= limit:
print "math.e**%s = %f" % (x, math.e**x)
print "last term = %s" % term
break
time.sleep(0.2)
n += 1
"""
Output for x == mpf(123.45):
sum =
410822093109668964239148908443317876138879647013995774.2951431466270782257597573259486687336246984942
k = 427
exp(123.45) =
410822093109668964239148908443317876138879647013995774.2951431466270782257597573259486687336246984942
e**123.45 =
410822093109668964239148908443317876138879647013995774.2951431466270782257597573259486687336246984942
"""
====================================================
This is also on the web at <http://python.pastebin.com/f1a5b9e03>.
Examples of problem x's: 10, 20, 30, 40, 100, 101
Examples of OK x's: 0.2, 5, 10.1, 11, 33.3, 123.45
Thanks,
Dick Moores
Dick Moores
>On Mon, 19 Nov 2007 23:41:02 -0800, Dick Moores <r...@rcblue.com>
>declaimed the following in comp.lang.python:
>
> > a way to get it to break where I want it to, i.e., when the sum
> > equals the limit as closely as the precision allows?
> >
>
>
> Well, since it ISN'T a case of testing for an absolute equivalence
>with floats...
>
> Perhaps putting a "print sum, limit" before that point would reveal
>what type of values you are encountering.
If you run the program you'll see exactly that, if I understand you
correctly. <http://python.pastebin.com/f2f06fd76> shows the full
output for a precision of 50 and x == 5.
Dick
Fredrik Johansson
> I'm writing a demo of the infinite series
>
> x**0/0! + x**1/1! + x**2/2! + x**3/3! + ... = e**x (x is non-negative)
>
> It works OK for many x, but for many the loop doesn't break. Is there
> equals the limit as closely as the precision allows?
>
Hi,
Checking that sum >= e**x will generally not work, because e**x might
have been rounded up while the sum might repeatedly be rounding down.
If this happens, no matter how many terms you add, the sum will never
reach the limit (one of the curses of finite-precision arithmetic).
One solution is to use directed rounding. First compute the limit with
downward rounding:
mpf.round_down()
limit = e**x
mpf.round_default()
Then compute every term in the sum with upward rounding:
mpf.round_down()
fac = factorial(n)
mpf.round_up()
term = x**n / fac
sum += term
(Note that the factorial should be rounded down to obtain upward
rounding in the term, since you're taking its reciprocal.)
This should guarantee that the sum eventually exceeds the limit.
As a simpler, less rigorous alternative, instead of checking if sum >=
limit, check (for example) whether
abs(sum - limit) / limit <= mpf(10)**(-precision+3)
i.e., if the sum is within 3 digits of the limit. This is the usual
way to test for numerical equality of floating-point numbers.
Fredrik
Dick Moores
I tried out both ways, and found that the second one best suited my
purposes. Please see the 2 highlighted lines in
<http://python.pastebin.com/fcc23b10>
Note that to break the loop I found that this does the job:
if abs(sum - limit) / limit <= mpf(10)**(-precision+1): # I
changed your +3 to +1
break
Fredrik, thanks VERY much for your terrific instruction!
Dick
davis...@gmail.com
for convergence? I.e., if sum == last_sum: break. Seems like that
would be more robust (you don't need to know the answer to computer
the answer), since it seems like it should converge.
--Nathan Davis
Dick Moores
>Instead of comparing sum to the "known" value of e**x, why not test
>for convergence? I.e., if sum == last_sum: break. Seems like that
>would be more robust (you don't need to know the answer to computer
>the answer), since it seems like it should converge.
Yes! And believe it or not I did that before seeing your post. Works
well. See <http://python.pastebin.com/f7c37186a>
And also with the amazing Chudnovsky algorithm for pi. See
<http://python.pastebin.com/f4410f3dc>
Thanks,
Dick
Fredrik Johansson
> And also with the amazing Chudnovsky algorithm for pi. See
> <http://python.pastebin.com/f4410f3dc>
Nice! I'd like to suggest two improvements for speed.
First, the Chudnovsky algorithm uses lots of factorials, and it's
rather inefficient to call mpmath's factorial function from scratch
each time. You could instead write a custom factorial function that
only uses multiplications and caches results, something like this:
cached_factorials = [mpf(1)]
def f(n):
n = int(n)
if n < len(cached_factorials):
return cached_factorials[n]
p = cached_factorials[-1]
for i in range(len(cached_factorials), n+1):
p *= i
cached_factorials.append(p)
return p
(In some future version of mpmath, the factorial function might be
optimized so that you won't have to do this.)
Second, to avoid unnecessary work, factor out the fractional power of
640320 that occurs in each term. That is, change the "denom =" line to
denom = (f(3*k) * ((f(k))**3) * (640320**(3*k)))
and then multiply it back in at the end:
print 1/(12*sum/640320**(mpf(3)/2))
With these changes, the time to compute 1,000 digits drops to only 0.05 seconds!
Further improvements are possible.
Fredrik
Steven D'Aprano
> Instead of comparing sum to the "known" value of e**x, why not test for
> convergence? I.e., if sum == last_sum: break. Seems like that would be
> more robust (you don't need to know the answer to computer the answer),
> since it seems like it should converge.
That will only work if you know that the terms in your sequence are
monotonically decreasing: that is, if each term is smaller than the last.
It also assumes that the terms decrease reasonably rapidly, and you want
the full precision available in floats. Are you sure your algorithm is
that precise? It's one thing to produce 16 decimal places in your result,
but if only the first 12 of them are meaningful, and the last four are
inaccurate, you might as well not bother with the extra work.
--
Steven.
Dick Moores
Fredrik,
I'm afraid I'm just too dumb to see how to use your first suggestion
of cached_factorials. Where do I put it and def()? Could you show me,
even on-line, what to do? <http://py77.python.pastebin.com/f48e4151c>
You (or anyone) can submit an amendment to my code using the textbox.
I did make the denom change, and see that it does improve the speed a bit.
Thanks,
Dick
John Machin
(1) Replace line 8 (from mpmath import factorial as f) with Fredrik's
code. (2) Test it to see that it gave the same answers as before ...
[time passes]
Wow! [w/o psyco] Pi to 1000 decimal places takes 13 seconds with
original code and 0.2 seconds with Fredrik's suggestion. 2000: 99
seconds -> 0.5 seconds.
Fredrik Johansson
> Fredrik,
>
> I'm afraid I'm just too dumb to see how to use your first suggestion
> of cached_factorials. Where do I put it and def()? Could you show me,
> even on-line, what to do? <http://py77.python.pastebin.com/f48e4151c>
>
> I did make the denom change, and see that it does improve the speed a bit.
I edited the pastebin code, see: http://py77.python.pastebin.com/m6b2b34b7
Fredrik
Dick Moores
Wow. your f() is ingenious, Frederik. Thanks very much.
Any more tricks up your sleeve? You did say a post or so ago,
"Further improvements are possible."
Dick
Fredrik Johansson
> Wow. your f() is ingenious, Frederik. Thanks very much.
>
> Any more tricks up your sleeve? You did say a post or so ago,
> "Further improvements are possible."
The next improvement would be to find a recurrence formula for the
terms instead of computing them directly. So for example, if summing
over c[n] = x**n / n!, instead of computing c[n] = x**n / factorial(n)
for each n, you'd compute c[0] and then just do c[n] = c[n-1] * x / n
for each of the remaining terms.
Fredrik