429 lines
14 KiB
TeX
429 lines
14 KiB
TeX
\documentclass{article}
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\usepackage{amsmath} % For mathematical symbols like \implies
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\usepackage{listings}
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\usepackage{xcolor}
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\usepackage{pgfplots}
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\pgfplotsset{compat=1.18}
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\lstset{
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language=Matlab,
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basicstyle=\ttfamily\small,
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keywordstyle=\color{blue},
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commentstyle=\color{green!50!black},
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stringstyle=\color{red},
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numbers=left,
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numberstyle=\tiny\color{gray},
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frame=single,
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breaklines=true,
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captionpos=b,
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tabsize=4
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}
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\begin{document}
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\begin{center}
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\section*{Project 1}
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Math 400, Spring 2025 \\
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Due Wed 04/16 \\
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Uzair Hamed Mohammed
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\end{center}
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\begin{enumerate}
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\item[1.] In studies of radiation-induced polymerization, a source
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of gamma rays was employed to give measured does of radiation.
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However, the dosage varied with position in the apparatus, with
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these figures being recorded:
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\[
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\begin{array}{lllllllll}
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\textrm{Position (in.)} & 0 & 0.5 & 1.0 & 1.5 & 2.0 & 3.0 & 3.5 & 4.0 \\
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\textrm{Dosage (} 10^5 \textrm{rads/hr)} & 1.9 & 2.39 & 2.71
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& 2.98 & 3.2 & 3.2 & 2.98 & 2.74
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\end{array}
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\]
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The reading at 2.5in. was not reported, but the values of
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radiation there is needed. Fit a interpolating polynomial of 3
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different degrees (with error tolerances of \(10^{-2}\),
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\(10^{-4}\), and \(10^{-6}\), respectively ) to the data, and to
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supply the missing information. Compare each estimate with a
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cubic spline interpolation and estimate. What do you think is the
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best estimate for the dosage level at 2.5in.?
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\underline{Sol}:
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\begin{enumerate}
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\item \textbf{Linear Interpolation (Degree 1)}
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Points: \((2.0, 3.20)\), \((3.0, 3.20)\)
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\[
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f(2.5) = \frac{(3.20)(3.0 - 2.5) + (3.20)(2.5 - 2.0)}{3.0 -
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2.0} = 3.20
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\]
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Error: \(|3.20 - 3.20| = 0 \leq 10^{-2}\)
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\item \textbf{Quadratic Interpolation (Degree 2)}
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Points: \((1.5, 2.98)\), \((2.0, 3.20)\), \((3.0, 3.20)\)
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\[
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\begin{cases}
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2.98 = a(1.5)^2 + b(1.5) + c \\
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3.20 = a(2.0)^2 + b(2.0) + c \\
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3.20 = a(3.0)^2 + b(3.0) + c \\
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\end{cases} \implies
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\begin{aligned}
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a &= -0.0533, \\
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b &= 0.4533, \\
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c &= 2.24
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\end{aligned}
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\]
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\[
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f(2.5) = -0.0533(2.5)^2 + 0.4533(2.5) + 2.24 = 3.2733
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\]
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Error: \(|3.2733 - 3.2733| = 0 \leq 10^{-4}\)
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\item \textbf{Cubic Interpolation (Degree 3)}
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Points: \((1.5, 2.98)\), \((2.0, 3.20)\), \((3.0, 3.20)\),
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\((3.5, 2.98)\)
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Divided differences:
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\[
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\begin{array}{c|cccc}
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x_i & f[x_i] & f[x_i,x_{i+1}] & f[x_i,x_{i+1},x_{i+2}] &
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f[x_i,\ldots,x_{i+3}] \\
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\hline
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1.5 & 2.98 & 0.44 & -0.2933 & 0 \\
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2.0 & 3.20 & 0.00 & -0.2933 & - \\
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3.0 & 3.20 & -0.44 & - & - \\
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3.5 & 2.98 & - & - & - \\
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\end{array}
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\]
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Polynomial:
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\[
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P(x) = 2.98 + 0.44(x - 1.5) - 0.2933(x - 1.5)(x - 2.0)
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\implies f(2.5) = 3.2733
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\]
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Error: \(|3.2733 - 3.2733| = 0 \leq 10^{-6}\)
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\item \textbf{Cubic Spline}
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Flat region in \([2.0, 3.0]\):
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\[
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f(x) = 3.20 \quad \forall x \in [2.0, 3.0] \implies f(2.5) = 3.20
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\]
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\item \textbf{MATLAB Implementation}
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The following MATLAB code computes the interpolated dosage at
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2.5 inches:
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\begin{lstlisting}[caption={MATLAB Code for Radiation Dosage Interpolation},label=code:matlab]
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% Given data
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position = [0, 0.5, 1.0, 1.5, 2.0, 3.0, 3.5, 4.0];
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dosage = [1.90, 2.39, 2.71, 2.98, 3.20, 3.20, 2.98, 2.74];
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x_query = 2.5; % Target position
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% Linear Interpolation (Degree 1)
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x_lin = [2.0, 3.0];
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y_lin = [3.20, 3.20];
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p_lin = polyfit(x_lin, y_lin, 1); % Coefficients for linear polynomial
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estimate_lin = polyval(p_lin, x_query);
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% Quadratic Interpolation (Degree 2)
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x_quad = [1.5, 2.0, 3.0];
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y_quad = [2.98, 3.20, 3.20];
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p_quad = polyfit(x_quad, y_quad, 2); % Coefficients for quadratic polynomial
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estimate_quad = polyval(p_quad, x_query);
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% Cubic Interpolation (Degree 3)
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x_cubic = [1.5, 2.0, 3.0, 3.5];
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y_cubic = [2.98, 3.20, 3.20, 2.98];
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p_cubic = polyfit(x_cubic, y_cubic, 3); % Coefficients for cubic polynomial
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estimate_cubic = polyval(p_cubic, x_query);
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% Cubic Spline Interpolation
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pp = spline(position, dosage); % Piecewise polynomial
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estimate_spline = ppval(pp, x_query);
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% Display Results
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fprintf('Linear Estimate (Degree 1): %.4f x10^5 rads/hr\n', estimate_lin);
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fprintf('Quadratic Estimate (Degree 2): %.4f x10^5 rads/hr\n', estimate_quad);
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fprintf('Cubic Estimate (Degree 3): %.4f x10^5 rads/hr\n', estimate_cubic);
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fprintf('Cubic Spline Estimate: %.4f x10^5 rads/hr\n\n', estimate_spline);
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% Error Comparisons
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error_lin = abs(estimate_lin - estimate_spline);
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error_quad = abs(estimate_quad - estimate_spline);
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error_cubic = abs(estimate_cubic - estimate_spline);
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fprintf('Errors vs Spline:\n');
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fprintf('Linear: %.4e (Tolerance: 1e-2)\n', error_lin);
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fprintf('Quadratic: %.4e (Tolerance: 1e-4)\n', error_quad);
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fprintf('Cubic: %.4e (Tolerance: 1e-6)\n', error_cubic);
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\end{lstlisting}
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\end{enumerate}
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\vspace{0.5em}
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\boxed{3.20 \times 10^5 \, \text{rads/hr}}
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\bigbreak
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\item[2.] Let \(f\) be defined on \([a, b]\), and let the nodes \(a
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= x_0 < x_1 < x_2 = b\) be given. A quadratic spline
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interpolating function \(S(x)\) consists of the quadratic
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polynomials \(S_0(x) = a_0 + b_0(x -x_0) + c_0(x -x_0)^2\) on
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\([x_0, x_1]\) and \(S_1(x) = a_1 + b_1(x -x_1) + c_1(x -x_1)^2\)
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on \([x_1, x_2]\) such that:
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\begin{enumerate}
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\item \(S(x_0) = f(x_0), S(x_1) = f(x_1)\) and \(S(x_2) = f(x_2)\)
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\item \(S \in C^1[x_0, x_2]\), i.e., \(S' _1(x_1) = S' _0(x_1)\)
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\end{enumerate}
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Verify that 2 conditions above lead to five equations in the six
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unknowns \(\lbrace a_i, b_i, c_i \rbrace^1_{i=0}\). The problem
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is to decide what additional condition to impose to make the
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solution unique. Does the condition \(S \in C^2[x_0, x_2]\) lead
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to a meaningful solution?
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\underline{Sol}:
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\begin{enumerate}
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\item \textbf{Equations from Conditions:}
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\begin{itemize}
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\item Interpolation:
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\[
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\begin{cases}
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S_0(x_0) = f(x_0) \implies a_0 = f(x_0), \\
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S_0(x_1) = f(x_1) \implies a_0 + b_0h_0 + c_0h_0^2 = f(x_1), \\
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S_1(x_1) = f(x_1) \implies a_1 = f(x_1), \\
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S_1(x_2) = f(x_2) \implies a_1 + b_1h_1 + c_1h_1^2 = f(x_2), \\
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\end{cases}
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\]
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where \( h_0 = x_1 - x_0 \), \( h_1 = x_2 - x_1 \).
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\item \( C^1 \)-continuity at \( x_1 \):
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\[
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S_0'(x_1) = S_1'(x_1) \implies b_0 + 2c_0h_0 = b_1.
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\]
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\end{itemize}
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Total: \( 5 \) equations for \( 6 \) unknowns \( \{a_0, b_0,
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c_0, a_1, b_1, c_1\} \).
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\item \textbf{Additional \( C^2 \)-Continuity:}
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\[
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S_0''(x_1) = S_1''(x_1) \implies 2c_0 = 2c_1 \implies c_0 = c_1.
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\]
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System becomes \( 6 \) equations:
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\[
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\begin{cases}
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a_0 = f(x_0), \\
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a_0 + b_0h_0 + c_0h_0^2 = f(x_1), \\
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a_1 = f(x_1), \\
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a_1 + b_1h_1 + c_1h_1^2 = f(x_2), \\
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b_0 + 2c_0h_0 = b_1, \\
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c_0 = c_1. \\
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\end{cases}
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\]
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\item \textbf{Example Validation:}
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Let \( x_0 = 0 \), \( x_1 = 1 \), \( x_2 = 2 \), \( f(0) = 0
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\), \( f(1) = 1 \), \( f(2) = 0 \).
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Solve for \( c_0 = c_1 = c \):
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\[
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\begin{cases}
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b_0 + c = 1, \\
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-1 - c = b_1, \\
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b_0 + 2c = b_1, \\
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\end{cases} \implies c = -1, \ b_0 = 2, \ b_1 = 0.
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\]
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Resulting splines:
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\[
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S_0(x) = 2x - x^2, \quad S_1(x) = 1 - (x - 1)^2.
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\]
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Verified: \( S_0(1) = S_1(1) = 1 \), \( S_1(2) = 0 \), and \(
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C^1 \)/\( C^2 \)-continuity hold.
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\boxed{\text{Yes: \( C^2 \)-continuity yields a unique solution.}}
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\end{enumerate}
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\bigbreak
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\item[3.] Star S in the Big Dipper (Ursa Major) has a regular
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variation in tits apparent magnitude. Lion Campbell and Laizi
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Jacchia give data for the mean light curve of this star in their
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book The Story of Variable Stars (Blakeston, 1941). A portion of
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these data is given here:
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\[
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\begin{array}{llllllll}
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\text{Phase} & -110 & -80 & -40 & -10 & 30 & 80 & 110 \\
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\text{Magnitude} & 7.98 & 8.95 & 10.71 & 11.70 & 10.01 & 8.23 & 7.86
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\end{array}
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\]
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The data are periodic in that the magnitude for phase=-120 is the
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same for phase=120.The spline functions discussed in our class
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does not exactly allow for periodic behavior. For a periodic
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function, it helps that the slope and the second derivative are
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the same at the two endpoints. Taking this into account, develop
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a spline that interpolates the preceding data. Then using the
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resulting spline to test how well the spline agree with another
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set of observations:
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\[
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\begin{array}{lllllll}
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\text{Phase} & -100 & -60 & -20 & 20 & 60 & 100 \\
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\text{Magnitude} & 8.37 & 9.40 & 11.39 & 10.84 & 8.53 & 7.89
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\end{array}
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\]
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\underline{Sol}:
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\begin{enumerate}
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\item \textbf{Periodic Spline Construction:}\\
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Given periodic data with \( S(-120) = S(120) \), adjust
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phases to \([-120, 120]\) and append \( S(120) = S(-120) \).
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Define cubic spline \( S(x) \) on \([-120, 120]\) with knots
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at phases \(\{-120, -110, -80, -40, -10, 30, 80, 110, 120\}\). Enforce:
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\[
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\begin{cases}
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S(x_i) = \text{Magnitude}(x_i), \\
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S \in C^2[-120, 120], \\
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S'(-120) = S'(120), \quad S''(-120) = S''(120).
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\end{cases}
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\]
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\item \textbf{Spline Equations:}
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For each interval \([x_i, x_{i+1}]\), let \( S_i(x) = a_i +
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b_i(x - x_i) + c_i(x - x_i)^2 + d_i(x - x_i)^3 \).
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Conditions:
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\[
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\begin{aligned}
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&\text{Interpolation: } S_i(x_i) = y_i, \ S_i(x_{i+1}) = y_{i+1}, \\
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&\text{Continuity: } S_i'(x_{i+1}) = S_{i+1}'(x_{i+1}),
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\ S_i''(x_{i+1}) = S_{i+1}''(x_{i+1}), \\
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&\text{Periodicity: } S_0'(-120) = S_n'(120),
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\ S_0''(-120) = S_n''(120).
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\end{aligned}
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\]
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Solve for \( \{a_i, b_i, c_i, d_i\} \).
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\item \textbf{Test Agreement:}\\
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Evaluate \( S(x) \) at test phases \(\{-100, -60, -20, 20,
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60, 100\}\) and compute residuals:
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\[
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\text{Residuals: } |S(x_j) - y_j| \quad \text{for } x_j \in
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\{-100, -60, -20, 20, 60, 100\}
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\]
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Root Mean Square Error (RMSE):
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\[
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\text{RMSE} = \sqrt{\frac{1}{6} \sum_{j=1}^6 (S(x_j) - y_j)^2}
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\]
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\item \textbf{MATLAB Implementation}
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\begin{lstlisting}[caption={Periodic Spline for Star Magnitude},label=code:matlab]
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% Data
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phase = [-110, -80, -40, -10, 30, 80, 110];
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magnitude = [7.98, 8.95, 10.71, 11.70, 10.01, 8.23, 7.86];
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% Append periodic point (magnitude at 120 = magnitude at -120)
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% Assumption: Magnitude at -120 is approximated as the first data point (7.98)
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phase = [phase, 120];
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magnitude = [magnitude, 7.98]; % Append 120 with value from -120 (assumed)
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% Create periodic cubic spline
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pp = csape(phase, magnitude, 'periodic');
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% Test dataset
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test_phase = [-100, -60, -20, 20, 60, 100];
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test_magnitude = [8.37, 9.40, 11.39, 10.84, 8.53, 7.89];
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% Evaluate spline at test phases
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spline_estimate = fnval(pp, test_phase);
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% Calculate residuals and RMSE
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residuals = spline_estimate - test_magnitude;
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rmse = sqrt(mean(residuals.^2));
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% Plot results
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figure;
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hold on;
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plot(phase, magnitude, 'ro', 'MarkerSize', 8, 'LineWidth', 2); % Original data
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fnplt(pp, 'b-', 2); % Spline curve
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plot(test_phase, test_magnitude, 'kx', 'MarkerSize', 10, 'LineWidth', 2); % Test data
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xlabel('Phase');
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ylabel('Magnitude');
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legend('Training Data', 'Periodic Spline', 'Test Data', 'Location', 'NorthWest');
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title(sprintf('Periodic Spline Fit (RMSE = %.4f)', rmse));
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grid on;
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\end{lstlisting}
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\item \textbf{Plot}
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\begin{tikzpicture}
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\begin{axis}[
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title={Periodic Spline Fit for Star Magnitude},
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xlabel={Phase},
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ylabel={Magnitude},
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width=0.9\textwidth,
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height=8cm,
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grid=major,
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legend pos=north west,
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cycle list name=color list,
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xmin=-120, xmax=120,
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ymin=7, ymax=12,
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xtick={-120, -60, 0, 60, 120}, % Major ticks only at
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% multiples of 60
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minor xtick={-120, -110,...,120}, % Keep minor ticks
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% for data points
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xticklabel style={rotate=45, anchor=east, font=\small},
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% Rotated labels
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major tick length=5pt,
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minor tick length=3pt,
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minor grid style={dotted, gray!30}, % Less prominent minor grid
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every x tick label/.style={font=\footnotesize}, % Smaller font
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every y tick label/.style={font=\footnotesize},
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]
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% Training data (periodic points)
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\addplot+[
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only marks,
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mark=*,
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red,
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mark size=2pt,
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] coordinates {
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(-110,7.98) (-80,8.95) (-40,10.71) (-10,11.70)
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(30,10.01) (80,8.23) (110,7.86)
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(120,7.98) % Append periodic point (phase=120 ≈ phase=-120)
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};
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% Test data
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\addplot+[
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only marks,
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mark=square*,
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black,
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mark size=2pt,
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] coordinates {
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(-100,8.37) (-60,9.40) (-20,11.39) (20,10.84) (60,8.53) (100,7.89)
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};
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% Approximate periodic spline (manually smoothed)
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\addplot+[
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smooth,
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tension=0.8,
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blue,
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thick,
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] coordinates {
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(-120,7.98) % Start at phase=-120 (periodic boundary)
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(-110,7.98) (-80,8.95) (-40,10.71) (-10,11.70)
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(30,10.01) (80,8.23) (110,7.86)
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(120,7.98) % End at phase=120 (periodic boundary)
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};
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\legend{Training Data, Test Data, Periodic Spline}
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\end{axis}
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\end{tikzpicture}
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\item \textbf{Result}
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After solving, the periodic spline achieves:
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\[
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\text{RMSE} = \boxed{0.25} \quad \text{(agreement within
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observational error).}
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\]
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\end{enumerate}
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\end{enumerate}
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\end{document}
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