pi

最近在玩2样东西，一样是go语言（golang），一样是Raspberry Pi（树莓派）。那么能在Raspberry上运行go语言程序吗？ Golang的文档上说是支持多平台的，其中也包括了ARM，看来有戏了，兴冲冲打开 http://code.google.com/p/go/downloads/list （ golang编译器的下载页面 ） ，却发现没有arm平台的二进制大包下载。所以arm平台下的需要下载源码自己编译。 Raspberry上编译前准备及配置 一。俗话说巧妇难为无米之炊，要编译源码先得下源码，下载后这里解压到 /home/pi/go 目录下。 二。确认需要的工具系统已经具备了：mercurial 、gcc 和 libc6-dev 。如果不确定有没有只需输入下列命令： sudo apt-get install -y mercurial gcc libc6-dev 以上命令实现安装3个必需的工具。 三。配置环境变量。因为第一点的时候我把下载下来的源码解压到/home/pi/go 目录下了，所以将 GOROOT 变量值设置为该路径： export GOROOT=/home/pi/go export GOARCH=arm 但其实不设置的话也是可以的，它默认是等一下命令行执行位置的父目录。 四。也是配置环境变量。 GOARM 变量可以不设置，默认值为：“6″

In R I have written this function ifun <- function(m) { o = c() for (k in 1:m) { o[k] = prod(1:k) / prod(2 * (1:k) + 1) } o_sum = 2 * (1 + sum(o)) # Final result print(o_sum) } This function approximates constant pi , however, after m > 50 the approximation gets stuck, i.e. the approximation is the same value and don't get better. How can I fix this? Thanks. Let's go inside: o <- numeric(100) for (k in 1:length(o)) { o[k] = prod(1:k) / prod(2 * (1:k) + 1) } o # [1] 3.333333e-01 1.333333e-01 5.714286e-02 2.539683e-02 1.154401e-02 # [6] 5.328005e-03 2.486402e-03 1.170072e-03 5.542445e-04 2

I am using Boost.Multiprecision for wrappers around the mpfr backend, and I am having some trouble creating pi (and e or any other math constant) to my desired precision. I feel like what I want to do should be possible, because of the use of Boost.Math for constants on a tutorial page for Boost.Multiprecision . In the tutorial, they use fixed-precision numbers of types such as cpp_dec_float_50 -- I want to do it with variable_precision mpfr_float . Check out the following code: #include <boost/multiprecision/mpfr.hpp> #include <boost/math/constants/constants.hpp> #include <iostream> ... int

This question already has an answer here: How do I print a double value with full precision using cout? 11 answers I have tried to use long double type in my program to print out more digits of pi. But it only shows 5 digits decimal. Here is my code. int main(int argc, char** argv) { long double pi_18 = acos(static_cast<long double>(-1)); cout << "pi to 18:" << pi_18 << endl; return 0; } and this is my output: pi to 18: 3.14159 How can I fix this problem? Like so: #include <iomanip> #include <iostream> std::cout << std::setw(15) << pi_18 << std::endl; The width modifier only affects the next

My school give me an assignment to calculate pi. The result should be : Question 4 Accuracy set at : 1000 term pi 1 4 100 3.13159 200 3.13659 300 3.13826 400 ... ... ... The result in my program : term pi 1 4 100 3 200 3 300 3 400 ... ... ... I guess that when I do (4 / denominator), the result will lose the decimal number although I have changed some declarations of data type from int to double. (Some websites tell me to do this.) Maybe I do it wrongly. How can I deal with this problem? The following is my program. #include <iostream> using namespace std; class Four { private: int

I have this random function to calculate pi Monte Carlo style : max=10000000; format long; in = 0; tic for k=1:max x = rand(); y = rand(); if sqrt(x^2 + y^2) < 1 in = in + 1; end end toc calc_pi = 4*in/max epsilon = abs(pi - calc_pi) calcPi(100000000); If I could iterate this 10e100 times, could this algorithm compete with the world record ? If so, how can I find the number of iteration that will give the Nth digit? This is a nice exercise for calculating pi, but it is probably a very inefficient one. Some remarks: My statistics are rusty before I had my coffee, but I guess the error scales

I spent some time yesterday writing the solution for this challenge published on Reddit , and was able to get through it without cheating, but I was left with a couple of questions. Reference material here . This is my code. (ns baking-pi.core (:import java.math.MathContext)) (defn modpow [n e m] (.modPow (biginteger n) (biginteger e) (biginteger m))) (defn div [top bot] (with-precision 34 :rounding HALF_EVEN (/ (bigdec top) (bigdec bot)))) (defn pow [n e] (.pow (bigdec n) (bigdec e) MathContext/DECIMAL128)) (defn round ([n] (.round (bigdec n) MathContext/DECIMAL128)) ([n & args] (->> [n args]