VB中定义Arccos函数

游侠无极限

我是这样的：
& d$ H' E1 ?* o+ d5 D7 @  fPrivate Function Arccos(x As Double, n As Long)
Dim i As Long, y As Double
For i = 1 To n
2 Q% x8 ?2 l; W    y = y + (-1 / Sqr(1 + (x / n * i) ^ 2)) * x / n
Next
8 |, J9 k5 n- c& Q) ?0 @Arccos = y
End Function

用的是Arccos的微分来算近似值，怎么不对？

游侠无极限

Const pi As Double = 3.14159265359
2 D  p( r! {- qPrivate Function Arccos(x As Double, n As Long) ' n 为函数精度，建议为100
    Select Case x
    Case 1
        Arccos = 0
    Case -1
/ g6 @  f$ r, |# n- q        Arccos = pi
    Case -1 To 1
3 g+ w- |9 f" G) N        Dim dx As Double, y As Double, i As Long
9 ]# _8 h: w6 i$ @: d  G+ h/ K' ^        dx = x / n
        For i = 1 To n
            y = y + (-1 / Sqr(1 - (dx * i + dx / 2) ^ 2)) * dx
        Next
        Arccos = y + pi / 2
    Case Else
& B9 c; w6 [* s4 k        '异常
    End Select
End Function
( q( Z+ Q. B8 B+ _# b3 v
( s2 _; {- j2 z/ G. g搞定了

煎饼

公式对就没有问题拉