Lagrange-Interpolationspolynom

Gegeben sind $n$ Punkte einer unbekannten Funktion $y = f(x)$
Gesucht ist das Interpolationspolynom $P_n(x)$, dass an den gegebenen Punkten mit $f(x)$ übereinstimmt.



In [1]:

    
# input given points
points = [
    (0, 2),
    (1, 2),
    (2, 1),
    (3, -1),
]
#



In [2]:

    
n = len(points)
X = [p[0] for p in points]
Y = [p[1] for p in points]

$$L_{n,k} = \frac{(x - x_0) \cdots (x - x_{k-1}) \cdot (x - x_{k+1}) \cdots (x - x_n)} {(x_k - x_0) \cdots (x_k - x_{k-1}) \cdot (x_k - x_{k+1}) \cdots (x_k - x_n)}$$



In [3]:

    
L = []
var('x')
for k in range(n):
    Lnk = (prod([(x - X[i]) for i in range(len(X)) if i != k])) / (prod([(X[k] - X[i]) for i in range(len(X)) if i != k]))
    L.append(Lnk.simplify_full())



In [4]:

    
for i, l in enumerate(L):
    show(LatexExpr(r"L_{{ {}, {} }} = ".format(n, i)), l)









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}L_{ 4, 0 } = -\frac{1}{6} \, x^{3} + x^{2} - \frac{11}{6} \, x + 1$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}L_{ 4, 1 } = \frac{1}{2} \, x^{3} - \frac{5}{2} \, x^{2} + 3 \, x$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}L_{ 4, 2 } = -\frac{1}{2} \, x^{3} + 2 \, x^{2} - \frac{3}{2} \, x$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}L_{ 4, 3 } = \frac{1}{6} \, x^{3} - \frac{1}{2} \, x^{2} + \frac{1}{3} \, x$

$$P_n(x) = \sum_{k=0}^{n} f(x_k) \cdot L_{n,k}$$



In [5]:

    
P = sum([Y[k] * L[k] for k in range(n)])
show(P.simplify_full())









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{2} \, x^{2} + \frac{1}{2} \, x + 2$



In [6]:

    
# some sanity checking
for k in range(n):
    yc = P.subs(x=X[k])
    yg = Y[k]
    assert yc == yg, "computed y {} != {} given y".format(yc, yg)



In [7]:

    
R = max(max((p[0], p[1])) for p in points) + 2
scatter_plot(points) + plot(P, xmax=R, ymax=R, ymin=-R, xmin=-R)









    Out[7]:



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