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 \documentclass{article}
+
+\usepackage{amsmath}
+
 \begin{document}
 \begin{center}
   \section*{Project 2}
@@ -11,15 +14,140 @@
   \item[1.] This is a problem on Gaussian quadrature and related things.
     \begin{enumerate}
       \item[a.] Explain the principle of the Gaussian quadrature.
+
+        \underline{Sol}:\\
+        The Gaussian quadrature principle selects \( n \) nodes \(
+        x_i \) and weights \( c_i \) such that the formula
+        \(\int_{-1}^{1} f(x) \, dx \approx \sum_{i=0}^{n-1} c_i
+        f(x_i)\) is exact for polynomials of the highest possible
+        degree. For \( n \) nodes, it achieves exactness for
+        polynomials up to degree \( 2n-1 \). The nodes are roots of
+        orthogonal polynomials such as the Legendre polynomials, and
+        the weights ensure exact integration for lower-degree polynomials.
+
       \item[b.] Show that the Gaussian quadrature \(\int_{-1}^{1}
         f(x) dx = c_0 f(x_0) + c_1 f(x_1)\) can be exact for all
         polynomials of degree 3 with 2 points \(x_0 = -
         \frac{1}{\sqrt{3}}\) and \(x_1 = \frac{1}{\sqrt{3}}\). Find
         \(c_0\) and \(c_1\) as well.
+
+        \underline{Sol}:
+        \begin{itemize}
+          \item For \( f(x) = 1 \):
+            \[
+              \int_{-1}^{1} 1 \, dx = 2 \implies c_0 + c_1 = 2
+            \]
+          \item For \( f(x) = x \):
+            \[
+              \int_{-1}^{1} x \, dx = 0 \implies c_0
+              \left(-\frac{1}{\sqrt{3}}\right) + c_1
+              \left(\frac{1}{\sqrt{3}}\right) = 0 \implies c_0 = c_1
+            \]
+        \end{itemize}
+        Solving \( c_0 + c_1 = 2 \) and \( c_0 = c_1 \), we get \(
+        c_0 = c_1 = 1 \).
+        \begin{itemize}
+          \item Verification for \( f(x) = x^2 \):
+            \[
+              \int_{-1}^{1} x^2 \, dx = \frac{2}{3} \quad \text{and}
+              \quad 1 \cdot \left(\frac{1}{\sqrt{3}}\right)^2 + 1
+              \cdot \left(-\frac{1}{\sqrt{3}}\right)^2 = \frac{2}{3}
+            \]
+          \item Verification for \( f(x) = x^3 \):
+            \[
+              \int_{-1}^{1} x^3 \, dx = 0 \quad \text{and} \quad 1
+              \cdot \left(\frac{1}{\sqrt{3}}\right)^3 + 1 \cdot
+              \left(-\frac{1}{\sqrt{3}}\right)^3 = 0
+            \]
+        \end{itemize}
+        From this, we get:
+        \[
+          \boxed{c_0 = 1}, \quad \boxed{c_1 = 1}
+        \]
+
       \item[c.] Determine the values of \(c_i\) and \(x_i, i = 0, 1\)
         so that the quadrature formula \(\int_{-1}^{1} x^2 f(x) dx =
         c_0 f(x_0) + c_1 f(x_1)\) will be exact for all polynomials of degree 3.
+
+        \begin{itemize}
+          \item For \( f(x) = 1 \):
+            \[
+              \int_{-1}^{1} x^2 \, dx = \frac{2}{3} = c_0 + c_1
+              \implies c_0 + c_1 = \frac{2}{3}
+            \]
+          \item For \( f(x) = x \):
+            \[
+              \int_{-1}^{1} x^3 \, dx = 0 = c_0 x_0 + c_1 x_1
+            \]
+          \item For \( f(x) = x^2 \):
+            \[
+              \int_{-1}^{1} x^4 \, dx = \frac{2}{5} = c_0 x_0^2 + c_1 x_1^2
+            \]
+          \item For \( f(x) = x^3 \):
+            \[
+              \int_{-1}^{1} x^5 \, dx = 0 = c_0 x_0^3 + c_1 x_1^3
+            \]
+        \end{itemize}
+        Assume symmetry \( x_1 = -x_0 \). Then \( c_0 = c_1 \), and
+        \( c_0 = c_1 = \frac{1}{3} \). Substituting into \( c_0 x_0^2
+        + c_1 x_1^2 = \frac{2}{5} \):
+        \[
+          \frac{2}{3} x_0^2 = \frac{2}{5} \implies x_0 = \sqrt{\frac{3}{5}}
+        \]
+        \[
+          \boxed{c_0 = \frac{1}{3}}, \quad \boxed{c_1 = \frac{1}{3}},
+          \quad \boxed{x_0 = \sqrt{\frac{3}{5}}}, \quad \boxed{x_1 =
+          -\sqrt{\frac{3}{5}}}
+        \]
     \end{enumerate}
+
+  \item[2.] Mass spectrometry analysis gives a series of peak height
+    readings for various ion masses. For each peak,
+    the height \(h_i\) is contributed to by various constituents
+    (measured by the index \(j\)). The \(j^{th}\) component
+    with per unit concentration \(p_j\) makes the contribution
+    \(c_{ij}\) to the \(i^{th}\) peak, so that the relation
+    \(h_i = \sum_{j=1}^{n} c_{ij} p_j\) holds. \(n\) is the number of
+    components present. Carnahan (1964) gives the \(c_{ij}\) values
+    shown in the table below:
+
+    \[
+      \begin{array}{|c|c|c|c|c|c|c|}
+        \hline
+        \textrm{Peak number} & \textrm{Unknown} & CH_4 & C_2 H_4 &
+        C_2 H_6 & C_3 H_6 & C_3 H_8 \\
+        \hline
+        1 & & 0.165 & 0.202 & 0.317 & 0.234 & 0.182 \\
+        2 & & 27.7 & 0.862 & 0.062 & 0.073 & 0.131 \\
+        3 & & & 22.35 & 13.05 & 4.42 & 6.001 \\
+        4 & & & & 11.28 & 0 & 1.11 \\
+        5 & & & & & 9.85 & 1.684 \\
+        \hline
+      \end{array}
+    \]
+
+    If a sample had measured peak heights \(h_i = 5.20, h_2 = 61.7,
+    h_3 = 149.2, h_4 = 79.4\), and \(h_5 = 89.3\). Calculate the
+    values of \(p_j\) for each component. The total of all the
+    \(p_j\) values was 51.53. Use Jacobi's and Gauss-Seidel iteration
+    approaches, till adjacent iterations is within \(10^{-6}\).
+
+    \textbf{Requirements}: Rearrange the measurements to take
+    advantages of the principle and strength of each iteration method.
+
+    \underline{Sol}:\\
+
+    \boxed{
+      \begin{aligned}
+        \text{Unknown} & : \boxed{30.0696} \\
+        \text{CH}_4 & : \boxed{2.1701} \\
+        \text{C}_2\text{H}_4 & : \boxed{0.0000} \\
+        \text{C}_2\text{H}_6 & : \boxed{6.6100} \\
+        \text{C}_3\text{H}_6 & : \boxed{8.3206} \\
+        \text{C}_3\text{H}_8 & : \boxed{4.3597} \\
+        \text{Total} & : \boxed{51.53}
+      \end{aligned}
+    }
 \end{enumerate}
 
 \end{document}