编辑: star薰衣草 | 2018-11-04 |
1 Volume 7, Number
1 Volume 7, Number
1 Volume 7, Number
1 March
200 March
200 March
200 March
2002 2
2 2 C C C C April
200 April
200 April
200 April
2002 2
2 2 对对对对数数数数表表表表的的的的构构构构造造造造李李李李永永永永隆隆隆隆Olympiad Corner The 32nd Austrian Mathematical Olympiad 2001.
Problem 1. Prove that ∑ = ? ? ? ? ? ? ? ?
2001 0
25 2
25 1 k k is an integer. ([x] denotes the largest integer less than or equal to x.) Problem 2. Determine all triples of positive real numbers x, y and z such that both x + y + z =
6 and = + + z y x
1 1
1 2 C xyz
4 hold. Problem 3. We are given a triangle ABC and its circumcircle with mid-point U and radius r. The tangent '
c of the circle with mid-point U and radius 2r is determined such that C lies between c = AB and '
c , and '
a and '
b are defined analogously, yielding the triangle '
'
'
C B A . Prove that the lines joining the mid-points of corresponding sides of ABC ? and '
'
'
C B A ? pass through a common point. (continued on page 4) Editors: 张百康(CHEUNG Pak-Hong), Munsang College, HK 高子眉(KO Tsz-Mei) 梁达荣(LEUNG Tat-Wing) 李健贤(LI Kin-Yin), Dept. of Math., HKUST 吴镜波(NG Keng-Po Roger), ITC, HKPU Artist: 杨秀英(YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST for general assistance. On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is May 15, 2002. For individual subscription for the next five issues for the 01-02 academic year, send us five stamped self-addressed envelopes. Send all correspondence to: Dr. Kin-Yin LI Department of Mathematics The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong Fax: (852)
2358 1643 Email: [email protected] 在现今计算工具发达的年代,要 找出如 ln2 这个对数值只需一指之劳 . 但是大家有没有想过,在以前计算机 尚未出现的时候,那些厚厚成书的对 数表是如何精确地构造出来的? 当然,在历史上曾出现很多不同的构造 方法,各有其所长,但亦各有其所限. 下面我们将会讨论一个比较有系统的 方法,它只需要用上一些基本的微积 分技巧,就能够有效地构造对数表到 任意的精确度. 首先注意,ln(xy) = ln x + ln y, 所 以我们只需求得所有质数 p 的对数值 便可以由此算得其他正整数的对数 值. 由ln(1 + t) 的微分运算和几何级 数公式直接可得 t t t t t t t t dt d n n n n + ? + ? + + ? + ? = + = + ? ?
1 )
1 ( )
1 (
1 1
1 )
1 ln(
1 1
3 2 运用微积分基本定理(亦即微分和积 分是两种互逆的运算) ,即得下式: dt t t n x x x x x dt t x x n n n n x ∫ + ? + ? + + ? + ? = ∫ + = + ?
0 1
4 3
2 0
1 )
1 ( )
1 (
4 3
2 1
1 )
1 ln( 能够对於所有正整数 n 皆成立. 现在 我们去估计上式中的积分余项的大 小. 设x ? x f x g (hence g(x) >
f(x)) for all x. Since sin x, f(x), g(x) ]
1 ,
1 [? ∈ ] , [
2 2 π π ? ? and sin x is strictly increasing in ] , [
2 2 π π ? , so f(x) is strictly increasing in ] , [
2 2 π π ? and ) ( ) ( ) ( x g g x g f x f f <