Kotlin使用BigDecimal与Double进行计算

我有两个功能。我们使用

biginger

和

BigDecimal

。我想用泰勒级数计算sin（z）：

这是我的密码：

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

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

根据谷歌的说法，Sin（1）是

0.8414709848

但是，以下内容的输出是：

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变量给出正确的值，而BigDecimal变量没有给出正确的值？即使有1000次迭代。

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注释掉的代码也用于计算Cos，但想先解决这个问题，所以我在

BigDecimal

变量中使两个函数看起来相同，尝试用

erg+=（zaehler/nenner）

替换

erg+=（zaehler.divide（nenner，20，RoundingMode.HALF\u偶数））

我怀疑缩放除法结果的默认值（如本文所述）不是您想要的

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顺便说一句-我假设性能不是练习的一部分，否则您的阶乘实现是一个低挂果实。

在

BigDecimal

变体中，尝试用

erg+=（zaehler/nenner）

替换

erg+=（zaehler.divide（nenner，20，RoundingMode.HALF_偶数））

我怀疑缩放除法结果的默认值（如本文所述）不是您想要的

---

---

顺便说一句-我假设性能不是练习的一部分，否则您的阶乘实现是一个很容易实现的结果。

我知道它的实现并不完美，甚至是罪恶。在性能方面，我有一个更好的实现，但我刚刚开始学习Kotlin，这只是练习语法的例子，与JavaI相比，我知道它的实现并不完美，甚至是Sin。在性能方面，我有一个更好的实现，但我刚刚开始学习Kotlin，这只是练习语法的例子，与Java相比