Kotlin 使用 BigDecimal 与 Double 进行计算

问题描述

我有两个功能。一种使用 BigInteger 和 BigDecimal。我想使用泰勒级数计算 sin(z)：

这是我的代码：

fun sinus(z: BigDecimal,upperBound: Int = 100): BigDecimal = calcSin(z,upperBound)

fun cosinus(z: BigDecimal,upperBound,false)

fun calcSin(z: BigDecimal,upperBound: Int = 100,isSin: Boolean = true): BigDecimal {
    var erg: BigDecimal = BigDecimal.ZERO
    for (n in 0..upperBound) {
//        val zaehler = (-1.0).pow(n).toBigDecimal() * z.pow(2 * n + (if (isSin) 1 else 0))
//        val nenner = fac(2 * n + (if (isSin) 1 else 0)).toBigDecimal()
        val zaehler = (-1.0).pow(n).toBigDecimal() * z.pow(2 * n + 1)
        val nenner = fac(2 * n + 1).toBigDecimal()
        erg += (zaehler / nenner)
    }
    return erg
}

fun calcSin(z: Double,upperBound: Int = 100): Double {
    var res = 0.0
    for (n in 0..upperBound) {
        val zaehler = (-1.0).pow(n) * z.pow(2 * n + 1)
        val nenner = fac(2 * n + 1,true)
        res += (zaehler / nenner)
    }
    return res
}

fun fac(n: Int): BigInteger = if (n == 0 || n == 1) BigInteger.ONE else n.toBigInteger() * fac(n - 1)

fun fac(n: Int,dummy: Boolean): Double = if (n == 0 || n == 1) 1.0 else n.todouble() * fac(n - 1,dummy)

根据 Google 的说法，Sin(1) 是

0.8414709848

然而，以下输出是：

println("Sinus 1: ${sinus(1.0.toBigDecimal())}")
println("Sinus 1: ${sinus(1.0.toBigDecimal()).todouble()}")
println("Sinus 1: ${sinus(1.0.toBigDecimal(),1000)}")
println("Sinus 1: ${calcSin(1.0)}")

输出：

Sinus 1: 0.8414373208078281027995610599000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Sinus 1: 0.8414373208078281
Sinus 1: 0.8414373208078281027995610599000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Sinus 1: 0.8414709848078965

我错过了什么？为什么 Double-Variant 给出了正确的值，而 BigDecimal 没有？即使有 1000 次迭代。 注释掉的代码也用于计算 Cos，但想先解决这个问题，所以我让两个函数看起来一样

解决方法

在 BigDecimal 变体中，尝试将 erg += (zaehler / nenner) 替换为 erg += (zaehler.divide(nenner,20,RoundingMode.HALF_EVEN))

我怀疑缩放除法结果的默认值（如此处 https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/math/BigDecimal.html 所述）不是您想要的。

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顺便说一句 - 我认为性能不是练习的一部分，否则你对阶乘的实现是一个低悬的果实。