查看“︁梅欽類公式”︁的源代码

'''梅钦类公式'''（英语：'''Machin-like formula'''）是[[数学]]中计算[[圆周率]]的一个常用技巧，它是[[約翰·梅欽|梅欽]]公式的推广，梅钦公式的形式为
:<math>\frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}</math>
梅钦依据此公式，把圆周率计算到一百多位小数。


梅钦类公式的形式为：
{{NumBlk|:|<math>c_0 \frac{\pi}{4} = \sum_{n=1}^N c_n \arctan \frac{a_n}{b_n}</math>|{{EquationRef|1}}}}

其中， <math>a_n</math> 和 <math>b_n</math> 为正 [[整数]]，且 <math>a_n < b_n</math>， <math>c_n</math> 为非零整数，且<math>c_0</math> 为正整数。

梅钦类公式的应用可结合[[反正切|反正切函数]]的[[泰勒级数]]展开：

{{NumBlk|:|<math>\arctan x = \sum^{\infin}_{n=0} \frac{(-1)^n}{2n+1} x^{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ...</math>|{{EquationRef|4}}}}
==导出==
根据[[三角恒等式#角的和差恒等式|角的和差公式]]，

:<math>\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta</math>
:<math>\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta</math>

若
<math> -\frac{\pi}{2} < \arctan \frac{a_1}{b_1} + \arctan \frac{a_2}{b_2} < \frac{\pi}{2}.</math>
有
{{NumBlk|:|<math>\arctan \frac{a_1}{b_1} + \arctan \frac{a_2}{b_2} = \arctan\frac{a_1 b_2 + a_2 b_1}{b_1 b_2 - a_1 a_2},</math>|{{EquationRef|2}}}}

反复应用这一方程，可得到所有的梅欽類公式，比如最初的梅欽公式：

:<math>2 \arctan \frac{1}{5}</math>

::<math>= \arctan \frac{1}{5} + \arctan \frac{1}{5}</math>

::<math>= \arctan \frac {1\times5 + 1\times5}{5\times5 - 1\times1}</math>

::<math>= \arctan \frac {10}{24}</math>

::<math>= \arctan \frac {5}{12}</math>

:<math>4 \arctan \frac{1}{5}</math>

::<math>= 2 \arctan \frac{1}{5} + 2 \arctan \frac{1}{5}</math>

::<math>= \arctan \frac{5}{12} + \arctan \frac{5}{12}</math>

::<math>= \arctan \frac{5\times12 + 5\times12}{12\times12 - 5\times5}</math>

::<math>= \arctan \frac{120}{119}</math>

:<math>4 \arctan \frac{1}{5} - \frac{\pi}{4}</math>

::<math>= 4 \arctan \frac{1}{5} - \arctan \frac{1}{1}</math>

::<math>= 4 \arctan \frac{1}{5} + \arctan \frac{-1}{1}</math>

::<math>= \arctan \frac{120}{119} + \arctan \frac{-1}{1}</math>

::<math>= \arctan \frac{120\times1 + (-1)\times119}{119\times1 - 120\times(-1)}</math>

::<math>= \arctan \frac{1}{239}</math>

:<math>\frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}</math>
<!--
An insightful way to visualize equation {{EquationNote|2}} is to picture what happens when two complex numbers are multiplied together:

:<math>(b_1 + a_1 i)*(b_2 + a_2 i)</math>
::<math>=b_1 b_2 + a_2 b_1 i + a_1 b_2 i - a_1 a_2</math>
{{NumBlk|::|<math>=(b_1 b_2 - a_1 a_2) + (a_1 b_2 + a_2 b_1) *  i</math>|{{EquationRef|3}}}}

The angle associated with a complex number <math>(b_n + a_n i)</math> is given by:

:<math>\arctan \frac {a_n}{b_n}</math>

Thus, in equation {{EquationNote|3}}, the angle associated with the product is:

:<math>\arctan \frac{a_1 b_2 + a_2 b_1}{b_1 b_2 - a_1 a_2}</math>

Note that this is the same expression as occurs in equation {{EquationNote|2}}.  Thus equation {{EquationNote|2}} can be interpreted as saying that the act of multiplying two complex numbers is equivalent to adding their associated angles (see [[Complex_number#Multiplication.2C_division_and_exponentiation_in_polar_form|multiplication of complex numbers]]).

The expression:

:<math>c_n \arctan \frac{a_n}{b_n}</math>

is the angle associated with:

:<math>(b_n + a_n i)^{c_n}</math>

Equation {{EquationNote|1}} can be re-written as:

:<math>k * (1 + i)^{c_0} = \prod_{n=1}^N (b_n + a_n i)^{c_n}</math>

Where <math>k</math> is an arbitrary constant that accounts for the difference in magnitude between the vectors on the two sides of the equation.  The magnitudes can be ignored, only the angles are significant.
 -->

==用梅钦公式编程计算圆周率(C++)==
<syntaxhighlight lang="C++">
#include<stdio.h>
#include<iostream>
using namespace std;
int main(void)
{   //本程序每四位数输出，如果请求计算的位数不是4的整数倍，最后输出可能会少1~3位数
	long a[2]={956,80},b[2]={57121,25},i=0,j,k,p,q,r,s=2,t,u,v,N,M=10000;
	printf("%9cMachin%6cpi=16arctan(1/5)-4arctan(1/239)\nPlease input a number.\n",32,32);
	cin>>N,N=N/4+3;
	long *pi=new long[N],*e=new long[N];
	while(i<N)pi[i++]=0;
	while(--s+1)
	{
		for(*e=a[k=s],i=N;--i;)e[i]=0;
		for(q=1;j=i-1,i<N;e[i]?0:++i,q+=2,k=!k)
			for(r=v=0;++j<N;pi[j]+=k?u:-u)u=(t=v*M+(e[j]=(p=r*M+e[j])/b[s]))/q,r=p%b[s],v=t%q;
	}
	while(--i)(pi[i]=(t=pi[i]+s)%M)<0?pi[i]+=M,s=t/M-1:s=t/M;
	for(cout<<"3.";++i<N-2;)printf("%04ld",pi[i]);
	delete []pi,delete []e,cin.ignore(),cin.ignore();
	return 0;
}
</syntaxhighlight>

== 參考文獻 ==
* {{MathWorld|urlname=Machin-LikeFormulas|title=Machin-like formulas}}
* [http://numbers.computation.free.fr/Constants/Pi/piclassic.html The constant π] {{Wayback|url=http://numbers.computation.free.fr/Constants/Pi/piclassic.html |date=20120204044407 }}
* [http://www.mathpages.com/home/kmath373.htm Machin's Merit] {{Wayback|url=http://www.mathpages.com/home/kmath373.htm |date=20101127043601 }} at MathPages
* [https://web.archive.org/web/20120119034334/http://milan.milanovic.org/math/english/pi/machin.html Archimedes' constant pi - Machin's formula] gives a proof for the John Machin`s formula
* GUIDORIZZI, Hamilton Luiz. ''Um Curso de Cálculo'', vol. 4. 3ª edição. Rio de Janeiro: Livros Técnicos e Científicos Editora, 1999.

[[Category:数学恒等式]]
[[Category:圆周率算法]]
[[Category:级数]]