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ネイピア数の実装 [C/C++]

Last updated at Posted at 2018-12-01

#ネイピア数

今回はC99とC++14で動作するネイピア数eを計算するプログラムを作成する。

##関数e1の実装

###元の式

$\displaystyle e = \lim_{t \to 0} \left(1 + t \right )^\frac{1}{t}$

###近似式

$\displaystyle t = \frac{1}{2^{51}}$

$\displaystyle e = \left(1 + t \right )^\frac{1}{t}$

##関数e2の実装

###元の式

$\displaystyle e = \sum_{n \to 0}^{\infty} \left(\frac{1}{n!} \right )$

###近似式

$\displaystyle e = 2+\sum_{n \to 2}^{18} \left(\frac{1}{n!} \right )$

#ソースコード

C99
#include <stdio.h>
#include <stdint.h>
#include <math.h>

typedef uint_fast64_t u64;

//lim t->0 (1+t)^(1/t)
static const double t = 1.0 / ((u64)1 << 51);
double e1() {
	return pow(1.0 + t, 1.0 / t);
}

//sigma n=0~18 (1/n!)
double e2() {
	double e = 2.0;
	u64 n;
	for (u64 i = (u64)2; i < (u64)18; ++i) {
		n = (u64)1;
		for (u64 j = (u64)2; j <= i; ++j) n *= j;
		e += (1.0 / n);
	}
	return e;
}
static const double e = e2();
static const double ee = 2.718281828459045;

int main() {
	printf("%.16f\n", e1());
	printf("%.16f\n", e2());
	printf("%.16f\n", e);
	printf("%.16f\n", ee);
	return 0;
}
C++14
#include <cstdio>
#include <cstdint>
#include <cmath>

namespace math {

	using u64 = uint_fast64_t;

	//lim t->0 (1+t)^(1/t)
	constexpr double t{ 1.0 / ((u64)1 << 51) };
	double e1(const double t_ = t) {
		return std::pow(1.0 + t_, 1.0 / t_);
	}

	//sigma n=0~18 (1/n!)
	constexpr double e2() {
		double e{ 2.0 };
		u64 n{};
		for (u64 i = (u64)2; i < (u64)18; ++i) {
			n = (u64)1;
			for (u64 j = (u64)2; j <= i; ++j) n *= j;
			e += (1.0 / n);
		}
		return e;
	}
	constexpr double e{ e2() };
	constexpr double ee{ 2.718281828459045 };

}

int main() {
	std::printf("%.16f\n", math::e1());
	std::printf("%.16f\n", math::e2());
	std::printf("%.16f\n", math::e);
	std::printf("%.16f\n", math::ee);
	return 0;
}
実行結果
2.7182818284590446
2.7182818284590455
2.7182818284590455
2.7182818284590451

##ソースコードのライセンス

These codes are licensed under CC0.
CC0

ソースコードは自由に使用してください。

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