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37 lines
1.4 KiB
TeX
37 lines
1.4 KiB
TeX
\section*{Aufgabe 3}
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\begin{table}[H]
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\begin{tabular}{l|l|l|l|l}
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$f_i$ & 8 & 3 & 4 & 8 \\ \hline
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$x_i$ & -1 & 0 & 1 & 3 \\
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\end{tabular}
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\end{table}
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\subsection*{Teilaufgabe i}
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\begin{align}
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p(x) = \sum_{i=0}^3 f_i \cdot L_i(x)
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\end{align}
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mit
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\begin{align}
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L_0(x) &= \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0 - x_1)(x_0-x_2)(x_0-x_3)}
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= \ldots = \frac{x^3 - 4x^2 + 3x}{-8} \\
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L_1(x) &= \frac{x^3 - 3x^2 - x + 3}{3} \\
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L_2(x) &= \frac{x^3 - 2x^2 - 3x}{-4} \\
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L_3(x) &= \frac{x^3 - x}{24}
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\end{align}
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\subsection*{Teilaufgabe ii}
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Anordnung der dividierten Differenzen im so genannten Differenzenschema:
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\begin{table}[H]
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\begin{tabular}{llll}
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$f[x_0]=f_0=8$ & ~ & ~ & ~ \\
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$f[x_1]= 3$ & $f[x_0,x_1] = \frac{f[x_0] - f[x_1]}{x_0-x_1} = -5$ & ~ & ~ \\
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$f[x_2] = 4$ & $1$ & $3$ & ~ \\
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$f[x_3] = 8$ & $2$ & $\frac{1}{3}$ & $- \frac{2}{3} $ \\
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\end{tabular}
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\end{table}
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Also:
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\begin{align}
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p(x) &= f[x_0] + f[x_0,x_1] \cdot (x-x_0) + f[x_0, x_1, x_2] \cdot (x-x_0) \cdot (x-x_1) \\ & + f[x_0, x_1, x_2, x_3] \cdot (x-x_0) \cdot (x-x_1) \cdot (x-x_2) \\
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&= 8 - 5 \cdot (x-x_0) + 3 \cdot (x-x_0) \cdot (x-x_1) \\ & - \frac{2}{3} \cdot (x-x_0) \cdot (x-x_1) \cdot (x-x_2)
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\end{align}
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