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tassadar
2019-06-08, 17:39
版面: 數學、科學、技術
主題: 圓週率公式(A): SUM(1/(2*n-1)^2) == ((π^2) / 8), n=1, 2, 3, 4, ...
回覆: 1
觀看: 66

Re: 圓週率公式(A): SUM(1/(2*n-1)^2) == ((π^2) / 8), n=1, 2, 3, 4, ...

PHP example of PI: 3.14159 3.14159 $ cat pi-odd-number.php <?php echo "scale=64; sqrt( 8 * ("; for ($i=1; $i<=16777215; $i+=2) { echo "1/($i*$i) + "; } echo " 0 )) "; ?> $ time php pi-odd-number.php | bc 3.1415926156442976203420241870879091412354242931426474819889608955 real 1m9.034s user 1m10.228s...
tassadar
2019-06-08, 16:43
版面: 數學、科學、技術
主題: 圓週率公式(A): SUM(1/(2*n-1)^2) == ((π^2) / 8), n=1, 2, 3, 4, ...
回覆: 1
觀看: 66

圓週率公式(A): SUM(1/(2*n-1)^2) == ((π^2) / 8), n=1, 2, 3, 4, ...

圓週率公式(A): SUM(1/(2*n-1)^2) == ((π^2) / 8), n=1, 2, 3, 4, ... 舉例: n = 1 ~ 512, 則 2^n-1 分母為積數 (Odd Number) 1, 3, 5, 7, 9 ... 至 1023 e.g. ----------------------------------------- SUM(1/(2*n-1)^2) == ((π^2) / 8) ----------------------------------------- sqrt ( 8 * (1/(1*1) + 1/(3*3) + 1/(5*5) + 1/(7*7...