无法让循环中断 [英] Trouble getting loop to break
问题描述
我正在写一个无限系列的演示
x ** 0/0！ + x ** 1/1！ + x ** 2/2！ + x ** 3/3！ + ... = e ** x（x是非负的）
对许多x都可以正常工作，但对于很多人而言，循环并没有中断。有没有
a的方式让它在我想要的地方打破，也就是说，当总和
等于限制时，精确度允许？
这就是我所拥有的：
======= series_xToN_OverFactorialN.py ============= =============
＃！/ usr / bin / env python
#coding = utf-8
＃series_xToN_OverFactorialN.py限制是来自p.63的e ** x
快乐的Pi，e
来自mpmath import mpf，e，exp，factorial
导入数学
导入时间
精度= 100
mpf.dps =精度
n = mpf（0）
x = mpf（raw_input（"输入非负int或浮点数：））
term = 1
sum = 0
limit = e ** x
k = 0
而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）
打印
如果总和> =限制：
print" math.e **％ s =％f ％（x，math.e ** x）
print" last term =％s" ％term
break
time.sleep（0.2）
n + = 1
" "
输出x == mpf（123.45）：
sum =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
K = 427
EXP（123.45）=
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
$ b $是** 123.45 =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
"""
============================ ====================== ==
这也是在网上的< http：// python .pastebin.com / f1a5b9e03> ;.
问题x的例子：10,20,30,40,100,101
OK的例子x'的：0.2,5,10.1,11,33.3,123.45
谢谢，
Dick Moores
2007年11月20日12:45，Dennis Lee Bieber写道：
> 2007年11月19日星期一23 ：41：02 -0800，Dick Moores< rd *@rcblue.com>
在comp.lang.python中声明了以下内容：
a to way让它打破我想要的地方，即，当总和
等于限制时，精确度允许？
如果总和> =限制：
嗯，因为它是ISN''用浮标测试绝对等价的情况......
或许放一个打印总和，限制在此之前会揭示您遇到的是什么类型的价值观。
如果你运行程序，你会看到，如果我理解你
正确。 < http://python.pastebin.com/f2f06fd76显示完整的
输出，精度为50，x == 5.
Dick
而不是将sum与已知相比较。 e ** x的值，为什么不测试
的收敛？即，如果sum == last_sum：break。好像那样
会更强大（你不需要知道电脑答案的答案），因为看起来它应该会收敛。 />
--Nathan Davis
11月20日凌晨1点41分，Dick Moores< r ... @ rcblue.comwrote：
我正在写一个无限系列的演示
x ** 0/0！ + x ** 1/1！ + x ** 2/2！ + x ** 3/3！ + ... = e ** x（x是非负的）
对许多x都可以正常工作，但对于很多人而言，循环并没有中断。有没有
a的方式让它在我想要的地方打破，也就是说，当总和
等于限制时，精确度允许？
这就是我所拥有的：
======= series_xToN_OverFactorialN.py ============= =============
＃！/ usr / bin / env python
#coding = utf-8
＃series_xToN_OverFactorialN.py限制是来自p.63的e ** x
快乐的Pi，e
来自mpmath import mpf，e，exp，factorial
导入数学
导入时间
精度= 100
mpf.dps =精度
n = mpf（0）
x = mpf（raw_input（"输入非负int或浮点数：））
term = 1
sum = 0
limit = e ** x
k = 0
而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）
打印
如果总和> =限制：
print" math.e **％ s =％f ％（x，math.e ** x）
print" last term =％s" ％term
break
time.sleep（0.2）
n + = 1
" "
输出x == mpf（123.45）：
sum =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
K = 427
EXP（123.45）=
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
$ b $是** 123.45 =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
"""
============================ ====================== ==
这也是在网上的< http：// python .pastebin.com / f1a5b9e03> ;.
问题x的例子：10,20,30,40,100,101
OK的例子x'的：0.2,5,10.1,11,33.3,123.45
谢谢，
Dick Moores
2007年11月20日上午10:42， da ***** ****@gmail.com 写道：
>而不是将总和与已知相比较。 e ** x的值，为什么不测试收敛？即，如果sum == last_sum：break。好像那样
会更健壮（你不需要知道电脑的答案答案），因为它似乎应该收敛。
是的！不管你信不信我在看到你的帖子之前就这么做了。工作
好。请参阅< http://python.pastebin.com/f7c37186a>
以及pi的惊人Chudnovsky算法。参见
< http://python.pastebin.com/f4410f3dc>
谢谢，
迪克
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
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 =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
k = 427
exp(123.45) =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
e**123.45 =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
"""
================================================== ==
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
At 12:45 AM 11/20/2007, Dennis Lee Bieber wrote:>On Mon, 19 Nov 2007 23:41:02 -0800, Dick Moores <rd*@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?
if sum >= limit:
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/f2f06fd76shows the full
output for a precision of 50 and x == 5.
Dick
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.
--Nathan Davis
On Nov 20, 1:41 am, Dick Moores <r...@rcblue.comwrote: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
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)
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 =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
k = 427
exp(123.45) =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
e**123.45 =
41082209310966896423914890844331787613887964701399 5774.295143146627078225759757325948668733624698494 2
"""
================================================== ==
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
At 10:42 AM 11/20/2007, da*********@gmail.com wrote:>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
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