prolog - 泰勒级数的近似值

Question

我有一些代码可以根据泰勒级数定义找到 sin(15°) 的近似值，并将其与内置函数 sin.

我有不同的结果。

• X - 弧度值
• R - 序列总和的当前值
• Eps - 精度值
• S - 符号
• F - N
的阶乘值
• XN - X 的 N 次方值
• N - 幂，1,3,5 ...

% base
sin_taylor(_,R,Eps,S,FN,XN,_) :-
    abs(S*XN/FN) < Eps,
    R = 0, !.

% step
sin_taylor(X, R, Eps, S, FN, XN, N) :-
    S1 = S*(-1),
    N1 = N+1,
    FN1 = FN*(2*N+2)*(2*N+3),
    XN1 = XN*X*X,
    sin_taylor(X,R1,Eps,S1,FN1,XN1,N1),
    R is R1+S*XN/FN.

% auxiliary predicate to supply parameters to the main one
sin(X,R,Eps) :-
    sin_taylor(X,R1,Eps,-1,1,X*X,1),
    R is 1+R1.

控制台中的结果:

?- X is 15*(pi / 180), sin(X,R,0.0001).
   X = 0.2617993877991494,
   R = 0.931695959721973.

?- X is 15*(pi / 180), R0 is sin(X).
   X = 0.2617993877991494,
   R0 = 0.25881904510252074.

Answer 1

无法识别代码中正确的第 n 项……阶乘计算隐藏在哪里？请记住，早期优化是软件工程中万恶之源:)

Wikipedia 中列出的解决方案的直接翻译引导我找到这段代码(好吧，应该在谓词名称中引用 Maclaurin 和 Taylor ...)

:- module(sin_taylor,
          [sin/3
          ,rad_deg/2
          ,fact/2
          ]).

% help predicate to give parameters to the main one
sin(X,R,Eps):-
    syn_taylor(X,Eps,0,R).

syn_taylor(X,Eps,N,R) :-
    S is -1**N,
    T is 2*N+1,
    fact(T,D),
    E is X**T,
    Q is S*E/D,
    (   Q < Eps
    ->  R = Q
    ;   M is N+1,
        syn_taylor(X,Eps,M,R1),
        R is Q+R1
    ).

rad_deg(R,D) :-
    var(R) -> R is D*(pi / 180).
    % tbd compute D from R

fact(N,F) :- N>0 -> N1 is N-1, fact(N1,F1), F is N*F1 ; F=1.

给出合理的结果:

?- rad_deg(X,15),sin(X,R,0.0001).
X = 0.2617993877991494,
R = 0.2588088132736575.

?- rad_deg(X,15),R0 is sin(X).
X = 0.2617993877991494,
R0 = 0.25881904510252074.