查看完整版本: 中译英，不知译得是否对，请诸位高手指正

中译英，不知译得是否对，请诸位高手指正

中译英（见附件），不知译得是否对，请诸位高手指正。
(考虑到想帮忙而下载还得扣书币，因此若想看本附件的网友可回复留Email,由我把附件发到邮箱。翻译是个很麻烦的事，看后不提意见也没关系。)给我留短信也可以。

请楼主不要在此版发附件贴。请直接把内容贴出来。附件需要下载，发Email也很麻烦耽误时间。如果真心请教，贴在这里很直观又省事

因数个数函数的推导证明

[quote]原帖由 [i]feuille1115[/i] 于 2008-3-19 20:57 发表 [url=http://www.chinesepdf.com/redirect.php?goto=findpost&pid=100188&ptid=50788][img]http://www.chinesepdf.com/images/common/back.gif[/img][/url]
请楼主不要在此版发附件贴。请直接把内容贴出来。附件需要下载，发Email也很麻烦耽误时间。如果真心请教，贴在这里很直观又省事 [/quote]
好的。只是因文中带有数学公式（是不必译的），在贴中不便显示，我就用
“数学公式（ x ）”表示，其中 x 为公式序号。


因数个数函数的推导证明
Induction and prove o Submultiples number function

包学行
Bao Xuexing

摘要　对自然数n的因数个数计数，把自然数n的因数个数记作F(n)，称为因数个数函数。本文证明了

数学公式()

Abstract  To count the number of submultiples of any natural number n, define number of submultiples of natural number n as F(n), and call it Submultiples Quantity Function(short to Sqf).  This essay has proved

数学公式()


关键词　自然数，因数，因数个数，因数个数函数。
Keyword　natural number, submultiples，number of submultiples，submultiples quantity function.



任一自然数n总有有限个可整除它的因数，设为F个，则F 与n之间存在一种函数关系。
For every natural number n, the number of its submultiples to aliquot it is limited. Define the number as  F(n) and there is a relation between  F and n which can be indicated by the function  
F(n).

设自然数n的能整除它的因数个数函数为F(n)，简称为“因数个数函数”。
Name the quantity of submultiples of natural number n as function F(n),and Sqf is short for Submultiples Quantity Function.

在一个纵坐标为F(t),横坐标为t的坐标平面上，我们可用在各区间[n - 0.5, n + 0.5 ]，上用一个脉宽为1，幅度为F(n)的单向矩形脉冲来表达“因数个数函数”的一种几何描述图形（图1上方）。
In an axis with y equal to F(t) and x equal to t, we can use a rectangle impulse with the width of 1 and range of F(n) in an interval[n-0.5,n+0.5] to indicate Submultiples Quantity Function.(As illustrated in diagram 1)

脉宽为1已把本来分布于各自然数点处的因数个数的值延伸给一个区间[kn - 0.5, kn+ 0.5 ]，因此自变量已由自然数n扩展为实数t。函数F(n)扩展为F(t), F(n)是F(t)中的一个“梳”,即
数学公式(1)

The width of 1 has extended the submultiples number of which here previously located on the dot of natural number to an interval of [Kn-0.5,Kn+0.5]. Therefore, the value extends from natural number n to primary number t, the function F(n) has extended to F(t). F(n) is a comb of F(t),that is to say,

数学公式(1)


图1   Diagram 1

再看周期为N，脉宽为1，幅度为1的单向矩形周期脉冲，各脉冲位于区间
[kN-0.5, kN+0.5]处，k为自然数。
Then look at the one-way rectangle impulsion with period of N, width of 1 and range of 1.  Every impulse is located inside the interval [Kn-0.5, Kn+0.5], in which k is a natural number.

这些脉冲的中心将位于所有能被N整除的数kN处。设这个周期脉冲为I(t,N)，则

数学公式(2)

The center of those impulsions will be located in the number Kn,which can be divided by N. Assume this impulsion I(t,N),then

数学公式(2)


(图1)下方画出了N=1,至 N=6的几个I(t,N) 的局部图形。（当N=1时前一脉冲的后沿与后一脉冲的前沿有一瞬间重叠，重叠的时间趋向于0，即无限短，本文予以忽略视为无重叠。因为即使不忽略，予以计算，对于本文的结果还是一样的。）
(Diagram 1) the following is partly the diagram of I(t,N) and from N=1 to N=6.（When N equals to 1，the previous impulse will overlap with the next impulse, the overlapping time is nearly zero. In this essay, we assume there is no overlapping, because even if we put the overlapping into calculation, the result is the same.)

我们知道波的迭加会产生“拍”，如果把 N=1,2,3,……,n  的所周期脉冲I(t, N)迭加，如果N自1至n 中有任何能整除n的数都将在区间[n-0.5，n+0.5]处有一个幅度为1的脉冲，这些脉冲的迭加将产生一个“拍”，“拍”的幅度值为能整除它的因数个数值，因此有

数学公式(3)

We know waves added together will produce beat. If we add up the impulse periods of N=1,2,3,...,n, and if there is any number from 1 to n to divide n, which will surely have a impulse of width 1  inside the interval of [n-0.5,n+0.5]. Those impulsions added together will produce a beat, and the range of the beat is the number of submultiples. Then we have a result as below

数学公式(3)

当N大于n时，首个脉冲已过于n之后了，对于t<=n+0.5部分是没有影响的，上式的作用相当于上述和式的上限为n=INT(t+0.5) ，下面再予讨论。
When N is larger than n，the impulse is behind n and won't affect the part when t is no larger than n +0.5, the equation above is mostly the upper finite of n =INT(t+0.5) . Here the discussion goes on.

I(N,t)的基频的角频率为
The angle frequency of basic frequency of I(t,N) is

数学公式(4)

我们可将I(t,N)展开为富里叶级数[1]
We can expand I(t,N) to Fourier Progression [1]

数学公式(5)

(5)式中的
in equation(5)

数学公式(6)
数学公式(7)
数学公式(8)

将(6)(7)(8)式代入(5)式，得
Put(6)(7)(8) into equation (5)，then

数学公式(9)


从图1 下方的I(t,N)图形可知，在区间(n-0.5, n+0.5)上I(t,N)为平行于t轴的水平线，在区间
(n-0.5, n+0.5)上对(2)式I(t,N)求关于t的导数，得

I’(t,N) = 0,                    (10)

We know from graph of I(t,N) below diagram(1)，in the interval(n-0.5, n+0.5), I(t,N) is a line parallel to t-axis, in the interval(n-0.5, n+0.5) according to equation(2)I(t,N) to figure out derivative of t，then

I’(t,N) = 0,                    (10)


根据李普希兹判别法则的推论[1]，因I(t,N)的导数在区间(n-0.5, n+0.5)上存在，所以在区间(n-0.5, n+0.5)上述(9)式收敛于I(t,N)。最关心的n点正在区间
(n-0.5, n+0.5)的中心。将(9)式代入(3)式，得
according to Liphiz Rule]，because derivative of I(t,N) is located between (n-0.5, n+0.5)，in the interval (n-0.5, n+0.5)the equation(9)is convergent on the point of I(t,N)。The most concerned dot n is in the center of (n-0.5, n+0.5)。Put equation (9) into equation (3)，then

数学公式(11)

在区间(n-0.5, n+0.5)上(11)式是n个收剑级数的和，上述(11)式收敛于F(t)。最关心的n点正在区间(n-0.5, n+0.5)的中心，因 F(n) 为F(t)在t=n时的一个“梳”，将t=n代入(11)式即得到因数个数函数可表达为

数学公式(12)

In the interval (n-0.5, n+0.5), equation(11) is to sum up convergent series ranged n，the equation (11) above is convergent on f(t). The most concerned dot n is located in the center of interval(n-0.5, n+0.5)，because  F(n) is a comb of F(t) when t is equal to n，put t=n into the equation(11)and have the result as

数学公式(12)


当N大于n时，首个脉冲已过于n之后了，对于t<=n+0.5部分是没有影响的，对于任一大于n的自然数u，有

数学公式(13)
数学公式(14)

When N is larger than n，the first impulse is behind n and has no effect on the part of  t<=n+0.5，for any natural number larger than u,

数学公式(13)
数学公式(14)


将(14)式代入(12)式，得因数个数函数为
Put equation (14) into equation (12)式，then the Submultiples Quantity Function is

数学公式()

因数个数函数的推导证明毕。
Then the induction of Submultiples Quantity Function completed.

————————————
参考文献：
Reference：
[1]数学分析（下册），复旦大学数学系主编，上海科学技术出版社1962年第二版，第721，741页。
[1]Mathematics Analysis，edited by Mathematics Department, Fudan University ,Science Technology Press, Shanghai, 1962,2nd Edition，P721，P741

   
    简介 本文定义F(n) 为n的因数个数函数，并证明了

数学公式()

SUMMARY　  Counting quantity of submultiple of natural number n , sign quantity of submultiple of n for F(n)，and called　a factor number function . This text proved

数学公式()
[/size]

————————————
[quote]因数学公式在贴中不便显示，如果有网友要看数学公式请发Email至[email]bsese@163.com[/email]索取“因数个数函数的推导证明.PDF”。[/quote]

试重译如下:
INDUCTION AND PROVING OF SUBMULTIPLES NUMBER FUNCTION
Xuexing Bao
Abstract: To count the number of submultiples of a natural number n, we define the number of submultiples of natural number n as F(n), and call it as Submultiples Quantity Function (or SQF).  In this paper we have proved
数学公式()
Keywords:　natural number, submultiples，number of submultiples，submultiples quantity function.
For any natural number n, there is a limited number of its submultiples to aliquot it. Define the number as  F(n) and there is a relationship between  F and n which can be indicated by the function F(n).
Name the quantity function of the submultiples of natural number n (the submultiples should be aliquot to n) as F(n), and call SQF as a short term of  Submultiples Quantity Function.
For an area with x as t and y as F(t), we may use a rectangle impulse with the width of 1 and amplitude of F(n) in an interval [n-0.5,n+0.5] to indicate SQF(shown in Fig. 1).
The width of 1 has extended the submultiples number of which here previously located on the dot of natural number to an interval of [Kn-0.5,Kn+0.5]. Therefore, the value has been extended from natural number n to primary number t. The function F(n) has been extended to F(t). F(n) is a comb of F(t), that is
Equation (1)
Figure 1
Then let's look at the one-way rectangle impulsion with period of N, width of 1 and amplitude of 1.  Every impulse is located inside the interval [Kn-0.5, Kn+0.5], where k is a natural number.
The center of those impulsions will locate at the number KN, which can be divided by N. Let this impulsion as I(t,N), we have
Equation (2)
The partition of I(t,N) with N from1 to 6 has been shown in Fig. 1. When N equals to 1，the previous impulse will overlap with the next impulse, the overlapping time is close to zero. In this paper, we assume there is no overlapping, because even if we take the overlapping into calculation, the result will be same.
We know that the superimposition of waves will produce beat. If we add up all impulses with the periods of N=1,2,3,...,n, and for any number from 1 to n to divide n, there must be an impulse with width of 1 inside the interval of [n-0.5,n+0.5]. The superimposition of those impulses will produce a beat. The amplitude of the beat is the number of submultiples. Therefore we have
Equation (3)
When N is larger than n，the impulse will pass n and not affect the part when t <=n +0.5, the equation (3) shows that the upper limit will be n =INT(t+0.5).  It will be discussed later.
The angle frequency of basic frequency of I(t,N) is
Equation (4)
I(t,N) may be expanded as Fourier series
Equation (5)
where,
Equation (6)
Equation (7)
Equation (8)
Substitute Eqs. (6), (7), and (8) to (5), we have
Equation (9)
From Fig. 1, we may see that in the interval (n-0.5, n+0.5) I(t,N) is a straight line which parallel to t-axis. In the interval(n-0.5, n+0.5) calculating the derivative of I(t,N) for t (according to equation 2), we have
I’(t,N) = 0                    (10)
According to Liphiz Rule [1], because derivative of I(t,N) exist in (n-0.5, n+0.5), therefore in the interval (n-0.5, n+0.5) the equation (9) is convergent on the point of I(t,N). The most concerned point n is right in the center of (n-0.5, n+0.5). Substitute equation (9) into (3), we have
Equation (11)
In the interval (n-0.5, n+0.5), equation (11) is the sum of n convergent series. It will convergent on F(t). The most concerned point n is located in the center of interval(n-0.5, n+0.5). Because  F(n) is a comb of F(t) when t is equal to n, substitute t=n into the equation (11) and we may have
Equation (12)
When N is larger than n, the first impulse is pass n and will not affect on the part of  t<=n+0.5. For any natural number larger than u, we have
Equation (13)
Equation (14)
Substitute equation (14) into (12)式. Submultiples Quantity Function (SQF) may be expressed as
Equation (15) ?
The induction of Submultiples Quantity Function completed.
----------------------
Reference:
[1] "Mathematics Analysis", Edited by Mathematics Department, Fudan University, Science Technology Press, Shanghai, 1962, 2nd Edition, p.721, p.741.
---------------------------------------
SUMMARY:　In this paper we defined F(n) as the Submultiples Quantity Function (SQF) of n and proved
Equation ()
===========================================
三点说明:
1. 因不知各公式的具体表述, 否则应译的更贴切.
2. 我的译文中指出有公式(15). 原文没有. 疑是有漏.
3. 如果方便的话, 请将原文(.PDF) Email to [email]Liberalist01@yahoo.com[/email]. 谢谢!

[quote]原帖由 [i]SH002[/i] 于 2008-3-22 18:55 发表 [url=http://www.chinesepdf.com/redirect.php?goto=findpost&pid=101108&ptid=50788][img]http://www.chinesepdf.com/images/common/back.gif[/img][/url]
试重译如下:
INDUCTION AND PROVING OF SUBMULTIPLES NUMBER FUNCTION
Xuexing Bao
Abstract: To count the number of submultiples of a natural number n, we define the number of submultiples of natural nu ... [/quote]
非常感谢[i]SH002[/i]的热情帮助。的确我张贴时漏了一个
数学公式()
这个公式在文首摘要时就已列出，所以就不标为数学公式(15)，而只标为
数学公式()
因数个数函数的推导证明(中英结照).pdf已发到你的邮箱，请注意查收。

Bsese,

参照你的PDF原文，我又作了少许修改。以下是修改过的段落。有什么问题，请告知（或Email给我）。相信我，你可试着投给你想投的杂志。
Good luck!

自由人
----------------------------------------------------------------------------
The center of those impulses will locate at the number KN, which may be divided by N. Take this impulsion as I(t,N), therefore
Equation (2)
      *      *       *      *
In the interval (n-0.5, n+0.5), equation (11) is the sum of n convergent series. It will convergent on F(t). The most concerned point n is located in the center of interval(n-0.5, n+0.5). Because  F(n) is a comb of F(t) when t is equal to n, substituting t=n into the equation (11), we have the Submultiples Quantity Function (SQF) F(n)
Equation (12)

[quote]原帖由 [i]SH002[/i] 于 2008-3-24 10:55 发表 [url=http://www.chinesepdf.com/redirect.php?goto=findpost&pid=101598&ptid=50788][img]http://www.chinesepdf.com/images/common/back.gif[/img][/url]
Bsese,

参照你的PDF原文，我又作了少许修改。以下是修改过的段落。有什么问题，请告知（或Email给我）。相信我，你可试着投给你想投的杂志。
Good luck!

自由人
------------------------------------------ ... [/quote]
非常感谢你的热心帮助，我确想将英译稿投稿，投稿时我必须要标明翻译者。在你之前的译文是[i]吴慧[/i]帮助翻译的，所以我想注明译者：[i]吴慧[/i]、[i]SH002[/i]（名）。不知你是否可Ｅmail告诉我你的实名。

呵，要实名的理由不够充分。如果杂志社一定要译者姓名，你注明是吴慧即可（她的确做了大量的工作）。
非常高兴能为我们村（地球村）做点什么。如果你找到一眼细流的涌泉为你的生命旅途解渴，用之，足矣。如果你因好奇而将泉眼凿大，泉废。

[quote]原帖由 [i]SH002[/i] 于 2008-3-24 14:49 发表 [url=http://www.chinesepdf.com/redirect.php?goto=findpost&pid=101648&ptid=50788][img]http://www.chinesepdf.com/images/common/back.gif[/img][/url]
呵，要实名的理由不够充分。如果杂志社一定要译者姓名，你注明是吴慧即可（她的确做了大量的工作）。
非常高兴能为我们村（地球村）做点什么。如果你找到一眼细流的涌泉为你的生命旅途解渴，用之，足矣。如果你因好 ... [/quote]
那么，只好在此先表示万分的感谢！

回复 8# 里“评分记录”的帖子

非常感谢管理员(feuille1115)的鼓励。望能在此多出力，广交友。

回复 8# 帖子

都是人生路人。无须言谢。互助共勉。吾村有望。（所言很“酸”，达意即可）

[quote]原帖由 [i]SH002[/i] 于 2008-3-25 12:18 发表 [url=http://www.chinesepdf.com/redirect.php?goto=findpost&pid=101928&ptid=50788][img]http://www.chinesepdf.com/images/common/back.gif[/img][/url]
非常感谢管理员(feuille1115)的鼓励。望能在此多出力，广交友。 [/quote]
非常高兴在此遇到了许多象 [i]SH002[/i] 那样热心的网友。

楼主太强悍了，一定加紧向楼主学习

[quote]原帖由 [i]hank1335[/i] 于 2008-3-29 04:24 发表 [url=http://www.chinesepdf.com/redirect.php?goto=findpost&pid=103354&ptid=50788][img]http://www.chinesepdf.com/images/common/back.gif[/img][/url]
楼主太强悍了，一定加紧向楼主学习 [/quote]
谢谢！强悍的不是楼主，而是象 [i]SH002[/i] 那样热心的网友、以及热心的管理员。还得感谢ChinesePDF提供了这样一个美好的环境。

太强了，遇到翻译就头疼

精辟！

[quote]原帖由 [i]SH002[/i] 于 2008-3-24 14:49 发表 [url=http://www.chinesepdf.com/redirect.php?goto=findpost&pid=101648&ptid=50788][img]http://www.chinesepdf.com/images/common/back.gif[/img][/url]
呵，要实名的理由不够充分。如果杂志社一定要译者姓名，你注明是吴慧即可（她的确做了大量的工作）。
非常高兴能为我们村（地球村）做点什么。如果你找到一眼细流的涌泉为你的生命旅途解渴，用之，足矣。如果你因好 ... [/quote]

比钱钟书的孔雀屁股的比喻雅多了。

回复 16# 的帖子

谢谢鼓励！
“物以类聚”。钱氏鉴于那个年代，有他们那一辈人的压抑和心酸。造就了那一代人的性格。但愿吾等之辈多些爱心和直率，少点冷酷和讥讽。

看看

我要，楼主我想看看，不知道可不可以？我的邮箱是[email]www80765869@126.com[/email]