拉马努金的几个神仙公式。据说他的论文里一共发表了14条圆周率的计算公式，但目前只收集到十个。
1 π = 1 8 ∑ m = 0 ∞ ( 20 m + 3 ) ( − 1 ) m ( 4 m ) ! ( 4 2 ) 4 m ( m ! ) 4 1 π = 1 2 3 ∑ m = 0 ∞ ( 8 m + 1 ) ( 4 m ) ! ( 4 3 ) 4 m ( m ! ) 4 1 π = 3 16 ∑ m = 0 ∞ ( 28 m + 3 ) ( − 1 ) m ( 4 m ) ! ( 64 3 ) 2 m ( m ! ) 4 1 π = 2 2 9 ∑ m = 0 ∞ ( 10 m + 1 ) ( 4 m ) ! 1 2 4 m ( m ! ) 4 1 π = 1 72 ∑ m = 0 ∞ ( 260 m + 23 ) ( − 1 ) m ( 4 m ) ! ( 12 2 ) 4 m ( m ! ) 4 1 π = 3 3 49 ∑ m = 0 ∞ ( 40 m + 3 ) ( 4 m ) ! 2 8 4 m ( m ! ) 4 1 π = 1 18 11 ∑ m = 0 ∞ ( 280 m + 19 ) ( 4 m ) ! ( 12 11 ) 4 m ( m ! ) 4 1 π = 5 288 ∑ m = 0 ∞ ( 644 m + 41 ) ( − 1 ) m ( 4 m ) ! ( 1152 5 ) 2 m ( m ! ) 4 1 π = 2 8 4 2 ∑ m = 0 ∞ ( 21460 m + 1123 ) ( − 1 ) m ( 4 m ) ! ( 84 2 ) 4 m ( m ! ) 4 1 π = 2 2 9 9 2 ∑ m = 0 ∞ ( 26390 m + 1103 ) ( 4 m ) ! 39 6 4 m ( m ! ) 4 begin{aligned} frac{1}{pi} &=frac{1}{8} sum^{infty}_{m=0}(20m+3)frac{(-1)^m(4m)!}{(4sqrt{2})^{4m}(m!)^4} & quad frac{1}{pi} &=frac{1}{2sqrt{3}} sum^{infty}_{m=0}(8m+1)frac{(4m)!}{(4sqrt{3})^{4m}(m!)^4} \ frac{1}{pi} &=frac{sqrt3}{16} sum^{infty}_{m=0}(28m+3)frac{(-1)^m(4m)!}{(64sqrt{3})^{2m}(m!)^4} & quad frac{1}{pi} &=frac{2sqrt{2}}{9} sum^{infty}_{m=0}(10m+1)frac{(4m)!}{12^{4m}(m!)^4} \ frac{1}{pi} &=frac{1}{72} sum^{infty}_{m=0}(260m+23)frac{(-1)^m(4m)!}{(12sqrt{2})^{4m}(m!)^4} & quad frac{1}{pi} &=frac{3sqrt3}{49} sum^{infty}_{m=0}(40m+3)frac{(4m)!}{28^{4m}(m!)^4} \ frac{1}{pi} &=frac{1}{18sqrt{11}} sum^{infty}_{m=0}(280m+19)frac{(4m)!}{(12sqrt{11})^{4m}(m!)^4} & quad frac{1}{pi} &=frac{sqrt{5}}{288} sum^{infty}_{m=0}(644m+41)frac{(-1)^m(4m)!}{(1152sqrt{5})^{2m}(m!)^4} \ frac{1}{pi} &=frac{2}{84^2} sum^{infty}_{m=0}(21460m+1123)frac{(-1)^m(4m)!}{(84sqrt{2})^{4m}(m!)^4} & quad frac{1}{pi} &=frac{2sqrt{2}}{99^2} sum^{infty}_{m=0}(26390m+1103)frac{(4m)!}{396^{4m}(m!)^4} end{aligned} π1π1π1π1π1=81m=0∑∞(20m+3)(42 )4m(m!)4(−1)m(4m)!=163 m=0∑∞(28m+3)(643 )2m(m!)4(−1)m(4m)!=721m=0∑∞(260m+23)(122 )4m(m!)4(−1)m(4m)!=1811 1m=0∑∞(280m+19)(1211 )4m(m!)4(4m)!=8422m=0∑∞(21460m+1123)(842 )4m(m!)4(−1)m(4m)!π1π1π1π1π1=23 1m=0∑∞(8m+1)(43 )4m(m!)4(4m)!=922 m=0∑∞(10m+1)124m(m!)4(4m)!=4933 m=0∑∞(40m+3)284m(m!)4(4m)!=2885 m=0∑∞(644m+41)(11525 )2m(m!)4(−1)m(4m)!=99222 m=0∑∞(26390m+1103)3964m(m!)4(4m)!
1989年，大卫·丘德诺夫斯基和格雷高里·丘德诺夫斯基兄弟将拉马努金公式改良。
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frac{1}{pi}=12sum^{infty}_{m=0} frac{545140134m+13591409}{640320^{3m+frac{3}{2}}} times frac{(-1)^m(6m)!}{(3m)!(m!)^3}
π1=12m=0∑∞6403203m+23545140134m+13591409×(3m)!(m!)3(−1)m(6m)!
百度拉马努金公式，搜索结果是最后一条，所以公式改良应该也是针对最后一条
用代码验证第一条公式：
我这里循环写到了26。因为循环在26到43之间取值都一样，毕竟笔记本精度有限；当超过43时，程序运行出错，毕竟笔记本能力有限
import math
sum = 0
for m in range(26):
x = (20*m+3)*((-1)**m)*math.factorial(4*m)
y = ((4*math.sqrt(2))**(4*m))*(math.factorial(m))**4
sum = sum + x/y
pi = 8/sum
# π = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510
print(pi)
print(math.pi)
'''
可以看到效果还是很不错的，起码比math库多一位，但真实值应该是2，准确度还差点。
3.1415926535897936
3.141592653589793
'''
改良公式：（改良还是有道理的，2到18取值一样，超过18，程序出错）
import math
sum = 0
for m in range(2):
x = (545140134*m+13591409)*((-1)**m)*math.factorial(6*m)
y = ((640320)**(3*m+1.5)) * (math.factorial(3*m)) * (math.factorial(m))**3
sum = sum + x/y
pi = 1/(12*sum)
# π = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510
print(pi)
print(math.pi)
'''
3.1415926535897936
3.141592653589793
'''
写了个Java代码，发现效果并不理想，循环超过10，输出就为NaN。
以下Java代码输出结果为：3.1463949385904666。而同是循环10次的python结果为：3.141593404158244
除了Math函数自身会将数值部分丢失，最大的问题就是Java的数据类型限制了范围。
public static void main(String[] args) {
Double sum = 0.0;
for (int m = 0; m < 10; m++) {
Double x = (20*m+3)*(Math.pow(-1, m))*factorial(4*m);
Double y = (Math.pow(4*Math.sqrt(2), 4*m))*power(factorial(m),4);
sum += x/y;
}
Double pi = 8/sum;
System.out.println(pi);
}
// 利用递归计算阶乘
public static int factorial(int num){
int sum = 1;
if(num==1 || num==0){
return 1;
}else{
sum=num * factorial(num-1);//运用递归计算
return sum;
}
}
// 计算次方
public static int power(int num, int n){
int pow = 1;
for (int i = 0; i < n; i++) {
pow *= num;
}
return pow;
}
