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\documentclass[uebung]{../../../lecture} |
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\title{Einführung in die Numerik: Übungsblatt 10} |
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\author{Leon Burgard, Christian Merten} |
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\begin{document} |
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\punkte |
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\begin{aufgabe} |
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Beh.: Seien $n$ paarweise verschiedene Stützstellen $\{x_0, \ldots, x_{n-1}\} $ gegeben |
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und eine Permutation derselben $\{\tilde{x}_0, \ldots, \tilde{x}_{n-1}\} $. Dann gilt |
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\[ |
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f[x_0, \ldots, x_{n-1}] = f[\tilde{x}_0, \ldots, \tilde{x}_{n-1}] |
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.\] |
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\begin{proof} |
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Es gilt nach VL mit der Newtondarstellung für das Interpolationspolynom |
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zu den Stützstellen $x_0, \ldots, x_{n-1}$: |
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\begin{salign*} |
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p(x) &= \sum_{i=0}^{n-1} y[x_0, \ldots, x_i] N_i(x) \\ |
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&= \sum_{i=0}^{n-2} y[x_0, \ldots, x_i] N_i(x) + y[x_0, \ldots, x_{n-1}] N_{n-1}(x) \\ |
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&= \mathcal{O}(x^{n-2}) + y[x_0, \ldots, x_{n-1}] \prod_{i=0}^{n-2} (x - x_i) \\ |
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&= \mathcal{O}(x^{n-2}) + y[x_0, \ldots, x_{n-1}] \left(x^{n-1} + \mathcal{O}(x^{n-2})\right) \\ |
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&= y[x_0, \ldots, x_{n-1}] x^{n-1} + \mathcal{O}(x^{n-2}) |
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.\end{salign*} |
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Der Leitkoeffizient des Interpolationspolynoms in der Monombasis ist also |
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$y[x_0, \ldots, x_{n-1}]$. Dieser ist unabhängig von der Reihenfolge der Stützstellen. Damit |
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folgt die Behauptung. |
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\end{proof} |
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\end{aufgabe} |
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\begin{aufgabe} |
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Beh.: Für $N \ge 2 \pi 10^{3}$ gilt $\displaystyle \max_{0 \le x \le 1} |f(x) - s(x)| < 10^{-12}$. |
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\begin{proof} |
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Es ist $f \in C^{4}([0, 1])$. Dann gilt nach VL |
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\[ |
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\delta \coloneqq \max_{0 \le x \le 1} |f(x) - s(x)| \le h^{4} \max_{0 \le x \le 1} |f^{(4)}(x)| |
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.\] Mit $f(x) = \sin(2\pi x)$ folgt sofort |
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\[ |
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f^{(4)}(x) = 16 \pi^{4} \sin(2 \pi x) \quad \text{also}\quad |
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\max_{0 \le x \le 1} |f^{(4)}(x)| = 16 \pi^{4} |
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.\] Mit $h = \frac{1}{N}$ ergibt sich |
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\[ |
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\delta \le 16 \frac{\pi^{4}}{N^{4}} \implies N \ge 2 \pi \sqrt[4]{\delta } = 2 \pi 10^{3} |
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.\] |
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\end{proof} |
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\end{aufgabe} |
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\begin{aufgabe} |
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Beh.: Das komplexe trigonometrische Interpolationspolynom ist gegeben als |
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\[ |
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t ^{*}(x) = \frac{1}{2} - \frac{1}{4} e^{ix} - \frac{1}{4} e^{3ix} |
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.\] |
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\begin{proof} |
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Die Stützstellen sind als $x_j = \frac{2 \pi j}{4}$, $j = 0, \ldots, 3$ gegeben. Damit folgt |
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\begin{salign*} |
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f(x_0) &= f(0) = \min \{0, 2\} = 0 \\ |
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f(x_1) &= f\left(\frac{\pi}{2}\right) = \min \left\{ \frac{1}{2}, \frac{3}{2} \right\} = \frac{1}{2} \\ |
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f(x_2) &= f(\pi) = \min \{1, 1\} = 1 \\ |
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f(x_3) &= f\left( \frac{3}{2}\pi \right) = \min \left\{ \frac{3}{2}, \frac{1}{2} \right\} = \frac{1}{2} |
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.\end{salign*} |
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Die Interpolationsbedingung ist erfüllt, denn |
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\begin{salign*} |
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t ^{*}(x_0) &= \frac{1}{2} - \frac{1}{4} - \frac{1}{4} = 0 = f(x_0) \\ |
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t ^{*}(x_1) &= \frac{1}{2} - \frac{1}{4} \underbrace{e^{i \frac{\pi}{2}}}_{= i} - \frac{1}{4} |
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\underbrace{e^{i \frac{3}{2} \pi}}_{= -i} = \frac{1}{2} = f(x_1) \\ |
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t ^{*}(x_2) &= \frac{1}{2} - \frac{1}{4} \underbrace{e^{i \pi}}_{= -1} - \frac{1}{4} |
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\underbrace{e^{i \pi}}_{= -1} = 1 = f(x_2) \\ |
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t ^{*}(x_3) &= \frac{1}{2} - \frac{1}{4} \underbrace{e^{i \frac{3}{2} \pi}}_{= i} - \frac{1}{4} |
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\underbrace{e^{i \frac{3}{2} \pi}}_{= -i} = \frac{1}{2} = f(x_3) |
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.\end{salign*} |
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Aus der Eindeutigkeit des komplexen trigonometrischen Interpolationspolynoms folgt die Behauptung. |
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\end{proof} |
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\end{aufgabe} |
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\begin{aufgabe} |
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\begin{enumerate}[a)] |
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\item |
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\begin{enumerate}[(1)] |
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\item Es ist |
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\[ |
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f_1''(x) = 6x - 14 \implies f_1''(0) = -14 \neq 0 |
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.\] Also erfüllt $f_1$ nicht die natürlichen Randbedingungen, also $f_1 \not\in S(X)$. |
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\item Es gilt für $0 \le x < 1$: |
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\begin{salign*} |
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f_2(x) &= -\frac{1}{2} x^{3} \xrightarrow{x \to 1} -\frac{1}{2}\\ |
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f_2'(x) &= -\frac{3}{2} x^2 \xrightarrow{x \to 1} - \frac{3}{2} \\ |
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f_2''(x) &= - 3x \xrightarrow{x \to 1} -3 \text{ und } f_2''(0) = 0 |
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\intertext{Für $1 \le x \le 2$ gilt:} |
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f_2(x) &= (x-1)^{3} -\frac{1}{2} x^{3} \implies f_2(1) = -\frac{1}{2} \\ |
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f_2'(x) &= 3(x-1)^2 - \frac{3}{2} x^2 \implies f_2'(1) = -\frac{3}{2} \\ |
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f_2''(x) &= 3x - 6 \implies f_2''(1) = -3 \text{ und } f_2''(2) = 0 |
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.\end{salign*} |
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$f_2$ ist auf beiden Teilintervallen ein Polynom von Grad $3$ und damit |
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auf den Teilintervallen beliebig oft stetig differenzierbar. Außerdem |
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ist $f_2$ $2$ mal stetig differenzierbar an der Stelle $1$, also insgesamt |
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$f_2 \in C^{2}([0, 2])$. Die natürlichen Randbedingungen sind außerdem erfüllt, also |
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folgt $f_2 \in S(X)$. |
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\item Es ist |
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\[ |
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f_3''(x) = 6x - 2 \implies f_3''(0) = -2 \neq 0 |
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.\] Also erfüllt $f_3$ nicht die natürlichen Randbedingungen, also $f_3 \not\in S(X)$. |
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\end{enumerate} |
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\item Der interpolierende Spline $s$ von $f(x) = x^{3}$ folgt mit der Darstellung der VL direkt |
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als |
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\[ |
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s(x) = \begin{cases} |
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1 + 4x + 4,5 x^2 + 1,5 x^{3} & x \in [0, 1) \\ |
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8 + 8,5(x-1) - 1,5(x-1)^{3} & x \in [1,2] |
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\end{cases} |
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.\] |
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\end{enumerate} |
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\end{aufgabe} |
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\begin{aufgabe} |
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Auszüge aus \textit{splines.cc}: |
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\begin{lstlisting}[language=C++, title=Schneller Löser für tridiagonale Matrizen, captionpos=b] |
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template<typename REAL> |
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void solveTriDiag(hdnum::DenseMatrix<REAL> &A, std::vector<REAL> &x, std::vector<REAL> &b) { |
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int N = b.size(); |
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x[0] = A[0][0]; |
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// LU Zerlegung |
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REAL l; |
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for (int j=1; j<N; j++) { |
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// berechne l faktor |
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l = A[j][j-1] / x[j-1]; |
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// modifiziere diagonalelemente und rechte seite |
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x[j] = A[j][j] - l * A[j-1][j]; |
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b[j] = b[j] - l * b[j-1]; |
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} |
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// Loesen durch Rueckwaertseinsetzen |
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x[N-1] = b[N-1] / x[N-1]; |
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for (int j=N-2; j>=0; j--) { |
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x[j] = (b[j] - A[j][j+1]*x[j+1])/x[j]; |
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} |
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}\end{lstlisting} |
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Implementation in einer Klasse \lstinline{CubicSpline}. Die Funktion \lstinline{getCubicSpline} |
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entspricht dem ersten Konstruktor. |
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\begin{lstlisting}[language=C++, title=Konstruktion und Auswertung eines kubischen Splines, captionpos=b] |
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template<typename REAL> |
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class CubicSpline { |
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public: |
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// erstelle einen kubischen spline mit vorgegebenen stuetzstellen |
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// und werten |
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CubicSpline(std::vector<REAL> xs, std::vector<REAL> ys) { |
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calculateCoefficients(xs, ys); |
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} |
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// erstelle einen kubischen spline mit vorgegebenen stuetzstellen |
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// und einer zu interpolierenden funktion |
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CubicSpline(std::vector<REAL> xs, REAL(*f)(REAL)) { |
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int N = xs.size(); |
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std::vector<double> ys(N); |
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for (int i=0; i<N; i++) { |
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ys[i] = f(xs[i]); |
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} |
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calculateCoefficients(xs, ys); |
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} |
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// erstelle einen kubischen spline an aequidistanten stuetzstellen |
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// und einer zu interpolierenden funktion |
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CubicSpline(REAL a, REAL b, int N, REAL(*f)(REAL)) { |
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std::vector<double> xs(N+1); |
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std::vector<double> ys(N+1); |
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for (int i=0; i<=N; i++) { |
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xs[i] = a + (1.0*i)/N*(b-a); |
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ys[i] = f(xs[i]); |
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} |
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calculateCoefficients(xs, ys); |
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} |
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// werte kubischen spline an vorgegebener stelle aus |
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REAL evaluate(REAL x) { |
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for (int i=1; i<x_s.size(); i++) { |
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if (x > x_s[i] && i < x_s.size() - 1) { |
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continue; |
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} else { |
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return a_0[i-1] + a_1[i-1] *(x - x_s[i]) + a_2[i-1]*std::pow(x - x_s[i], 2) + a_3[i-1]*std::pow(x-x_s[i], 3); |
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} |
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} |
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return 0; |
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} |
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// gebe alle interpolations polynome aus |
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void print() { |
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for(int i=0; i<x_s.size()-1; i++) { |
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printf("p_%d(x) = %4.2f + %4.2f(x - %4.2f) + %4.2f(x - %4.2f)^2 + %4.2f(x - %4.2f)^3\n", |
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i, a_0[i], a_1[i], x_s[i], a_2[i], x_s[i], a_3[i], x_s[i]); |
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} |
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} |
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private: |
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// stuetzstellen |
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std::vector<REAL> x_s; |
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// koeffizienten |
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std::vector<REAL> a_0; |
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std::vector<REAL> a_1; |
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std::vector<REAL> a_2; |
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std::vector<REAL> a_3; |
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void calculateCoefficients(std::vector<REAL> xs, std::vector<REAL> ys) { |
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// stuetzstellen from x_0 ... to x_n |
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int n = xs.size()-1; |
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// copy stuetzstellen |
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x_s = std::vector<double>(n+1); |
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x_s = xs; |
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// setup (n-1)x(n-1) matrix for a_2 |
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hdnum::DenseMatrix<REAL> A(n-1,n-1); |
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std::vector<REAL> b(n-1); |
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std::vector<REAL> x(n-1); |
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REAL h; // h_i |
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REAL h1; // h_{i+1} |
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// setup LGS for a_2 |
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// A has tridiagonal structure |
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for (int i = 1; i<n; i++) { |
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h = xs[i] - xs[i-1]; |
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h1 = xs[i+1] - xs[i]; |
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b[i-1] = 3 * ((ys[i+1] - ys[i])/h1 - (ys[i] - ys[i-1])/h); |
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if (i > 1) { |
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A[i-1][i-2] = h; |
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} if (i < n-1) { |
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A[i-1][i] = h1; |
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} |
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A[i-1][i-1] = 2 * (h + h1); |
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} |
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// initialize vectors |
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a_0 = std::vector<double>(n); |
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a_1 = std::vector<double>(n); |
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a_2 = std::vector<double>(n); |
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a_3 = std::vector<double>(n); |
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solveTriDiag(A,a_2,b); |
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// natuerliche randbedingung |
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a_2[n-1] = 0; |
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// berechne restliche koeffizienten |
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for (int i = 1; i<=n; i++) { |
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h = xs[i] - xs[i-1]; // h_i |
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h1 = xs[i+1] - xs[i]; // h_{i+1} |
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a_0[i-1] = ys[i]; |
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if (i == 1) { // a_2[-1] = 0 |
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a_1[i-1] = (ys[i] - ys[i-1])/h + (h/3)*(2*a_2[i-1]); |
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a_3[i-1] = (a_2[i-1])/(3*h); |
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} else { |
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a_1[i-1] = (ys[i] - ys[i-1])/h + h/3*(2*a_2[i-1] + a_2[i-2]); |
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a_3[i-1] = (a_2[i-1] - a_2[i-2])/(3 * h); |
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} |
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} |
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} |
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};\end{lstlisting} |
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\begin{figure}[h] |
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\centering |
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\begin{tikzpicture} |
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\begin{axis}[xtick=\empty, ytick=\empty] |
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\addplot[purple] table {saurier.dat}; |
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\end{axis} |
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\end{tikzpicture} |
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\caption{Rekonstruktion des Sauriers} |
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\label{fig:} |
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\end{figure} |
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\end{aufgabe} |
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\end{document} |
