Finished project 2
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project2.tex
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project2.tex
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\documentclass{article}
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\usepackage{amsmath}
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\begin{document}
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\begin{center}
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\section*{Project 2}
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\item[1.] This is a problem on Gaussian quadrature and related things.
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\begin{enumerate}
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\item[a.] Explain the principle of the Gaussian quadrature.
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\underline{Sol}:\\
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The Gaussian quadrature principle selects \( n \) nodes \(
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x_i \) and weights \( c_i \) such that the formula
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\(\int_{-1}^{1} f(x) \, dx \approx \sum_{i=0}^{n-1} c_i
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f(x_i)\) is exact for polynomials of the highest possible
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degree. For \( n \) nodes, it achieves exactness for
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polynomials up to degree \( 2n-1 \). The nodes are roots of
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orthogonal polynomials such as the Legendre polynomials, and
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the weights ensure exact integration for lower-degree polynomials.
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\item[b.] Show that the Gaussian quadrature \(\int_{-1}^{1}
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f(x) dx = c_0 f(x_0) + c_1 f(x_1)\) can be exact for all
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polynomials of degree 3 with 2 points \(x_0 = -
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\frac{1}{\sqrt{3}}\) and \(x_1 = \frac{1}{\sqrt{3}}\). Find
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\(c_0\) and \(c_1\) as well.
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\underline{Sol}:
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\begin{itemize}
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\item For \( f(x) = 1 \):
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\[
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\int_{-1}^{1} 1 \, dx = 2 \implies c_0 + c_1 = 2
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\]
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\item For \( f(x) = x \):
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\[
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\int_{-1}^{1} x \, dx = 0 \implies c_0
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\left(-\frac{1}{\sqrt{3}}\right) + c_1
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\left(\frac{1}{\sqrt{3}}\right) = 0 \implies c_0 = c_1
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\]
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\end{itemize}
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Solving \( c_0 + c_1 = 2 \) and \( c_0 = c_1 \), we get \(
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c_0 = c_1 = 1 \).
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\begin{itemize}
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\item Verification for \( f(x) = x^2 \):
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\[
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\int_{-1}^{1} x^2 \, dx = \frac{2}{3} \quad \text{and}
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\quad 1 \cdot \left(\frac{1}{\sqrt{3}}\right)^2 + 1
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\cdot \left(-\frac{1}{\sqrt{3}}\right)^2 = \frac{2}{3}
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\]
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\item Verification for \( f(x) = x^3 \):
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\[
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\int_{-1}^{1} x^3 \, dx = 0 \quad \text{and} \quad 1
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\cdot \left(\frac{1}{\sqrt{3}}\right)^3 + 1 \cdot
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\left(-\frac{1}{\sqrt{3}}\right)^3 = 0
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\]
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\end{itemize}
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From this, we get:
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\[
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\boxed{c_0 = 1}, \quad \boxed{c_1 = 1}
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\]
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\item[c.] Determine the values of \(c_i\) and \(x_i, i = 0, 1\)
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so that the quadrature formula \(\int_{-1}^{1} x^2 f(x) dx =
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c_0 f(x_0) + c_1 f(x_1)\) will be exact for all polynomials of degree 3.
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\begin{itemize}
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\item For \( f(x) = 1 \):
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\[
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\int_{-1}^{1} x^2 \, dx = \frac{2}{3} = c_0 + c_1
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\implies c_0 + c_1 = \frac{2}{3}
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\]
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\item For \( f(x) = x \):
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\[
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\int_{-1}^{1} x^3 \, dx = 0 = c_0 x_0 + c_1 x_1
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\]
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\item For \( f(x) = x^2 \):
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\[
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\int_{-1}^{1} x^4 \, dx = \frac{2}{5} = c_0 x_0^2 + c_1 x_1^2
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\]
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\item For \( f(x) = x^3 \):
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\[
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\int_{-1}^{1} x^5 \, dx = 0 = c_0 x_0^3 + c_1 x_1^3
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\]
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\end{itemize}
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Assume symmetry \( x_1 = -x_0 \). Then \( c_0 = c_1 \), and
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\( c_0 = c_1 = \frac{1}{3} \). Substituting into \( c_0 x_0^2
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+ c_1 x_1^2 = \frac{2}{5} \):
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\[
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\frac{2}{3} x_0^2 = \frac{2}{5} \implies x_0 = \sqrt{\frac{3}{5}}
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\]
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\[
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\boxed{c_0 = \frac{1}{3}}, \quad \boxed{c_1 = \frac{1}{3}},
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\quad \boxed{x_0 = \sqrt{\frac{3}{5}}}, \quad \boxed{x_1 =
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-\sqrt{\frac{3}{5}}}
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\]
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\end{enumerate}
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\item[2.] Mass spectrometry analysis gives a series of peak height
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readings for various ion masses. For each peak,
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the height \(h_i\) is contributed to by various constituents
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(measured by the index \(j\)). The \(j^{th}\) component
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with per unit concentration \(p_j\) makes the contribution
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\(c_{ij}\) to the \(i^{th}\) peak, so that the relation
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\(h_i = \sum_{j=1}^{n} c_{ij} p_j\) holds. \(n\) is the number of
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components present. Carnahan (1964) gives the \(c_{ij}\) values
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shown in the table below:
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\[
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\begin{array}{|c|c|c|c|c|c|c|}
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\hline
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\textrm{Peak number} & \textrm{Unknown} & CH_4 & C_2 H_4 &
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C_2 H_6 & C_3 H_6 & C_3 H_8 \\
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\hline
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1 & & 0.165 & 0.202 & 0.317 & 0.234 & 0.182 \\
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2 & & 27.7 & 0.862 & 0.062 & 0.073 & 0.131 \\
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3 & & & 22.35 & 13.05 & 4.42 & 6.001 \\
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4 & & & & 11.28 & 0 & 1.11 \\
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5 & & & & & 9.85 & 1.684 \\
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\hline
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\end{array}
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\]
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If a sample had measured peak heights \(h_i = 5.20, h_2 = 61.7,
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h_3 = 149.2, h_4 = 79.4\), and \(h_5 = 89.3\). Calculate the
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values of \(p_j\) for each component. The total of all the
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\(p_j\) values was 51.53. Use Jacobi's and Gauss-Seidel iteration
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approaches, till adjacent iterations is within \(10^{-6}\).
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\textbf{Requirements}: Rearrange the measurements to take
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advantages of the principle and strength of each iteration method.
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\underline{Sol}:\\
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\boxed{
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\begin{aligned}
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\text{Unknown} & : \boxed{30.0696} \\
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\text{CH}_4 & : \boxed{2.1701} \\
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\text{C}_2\text{H}_4 & : \boxed{0.0000} \\
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\text{C}_2\text{H}_6 & : \boxed{6.6100} \\
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\text{C}_3\text{H}_6 & : \boxed{8.3206} \\
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\text{C}_3\text{H}_8 & : \boxed{4.3597} \\
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\text{Total} & : \boxed{51.53}
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\end{aligned}
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}
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\end{enumerate}
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\end{document}
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