My name is Hanzhang Yin, and I have developed this website as a resource to facilitate the review of key concepts in abstract algebra, and it is not fully completed. Feel free to email me at hanyin@ku.edu if there are any errors or suggestions for improvement.

Solution (a). Gaussian quadrature is a method to approximate the integral of a function by evaluating the function at specific points and weighting the function values. The weights and points are chosen such that the method is exact for polynomials of a certain degree. \[ \int_a^b f(x)w(x)dx \approx \sum_{i=1}^n w_i f(x_i). \]

Solution (b). Set \(f(x) = 1\), we have \[ \begin{align*} I(f) &= \int_{-1}^1 1 \, dx = 2, \\ G_2(f) &= w_1 f \left( -\frac{1}{\sqrt{3}} \right) + w_2 f \left( \frac{1}{\sqrt{3}} \right) = w_1 + w_2 = 2. \end{align*} \] Now, set \(f(x) = x\), we have \[ \begin{align*} I(f) &= \int_{-1}^1 x \, dx = 0, \\ G_2(f) &= w_1 f \left( -\frac{1}{\sqrt{3}} \right) + w_2 f \left( \frac{1}{\sqrt{3}} \right) = w_1 \left( -\frac{1}{\sqrt{3}} \right) + w_2 \left( \frac{1}{\sqrt{3}} \right) = 0. \end{align*} \] Hence, we can get \(w_1 = w_2 = 1\). Now, set \(f(x) = x^2\), we have \[ \begin{align*} I(f) &= \int_{-1}^1 x^2 \, dx = \frac{2}{3}, \\ G_2(f) &= f \left( -\frac{1}{\sqrt{3}} \right) + f \left( \frac{1}{\sqrt{3}} \right) = \left( -\frac{1}{\sqrt{3}} \right)^2 + \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{2}{3}. \end{align*} \] Hence, we can know that the approximation is exact when the degree is 2. Then, we set \(f(x) = x^3\), we have \[ \begin{align*} I(f) &= \int_{-1}^1 x^3 \, dx = 0, \\ G_2(f) &= f \left( -\frac{1}{\sqrt{3}} \right) + f \left( \frac{1}{\sqrt{3}} \right) = \left( -\frac{1}{\sqrt{3}} \right)^3 + \left( \frac{1}{\sqrt{3}} \right)^3 = 0. \end{align*} \] Thus, we again have the approximation is exact when the degree is \(3\). After that, we set \(f(x) = x^4\), we have \[ \begin{align*} I(f) &= \int_{-1}^1 x^4 \, dx = \frac{2}{5}, \\ G_2(f) &= f \left( -\frac{1}{\sqrt{3}} \right) + f \left( \frac{1}{\sqrt{3}} \right) = \left( -\frac{1}{\sqrt{3}} \right)^4 + \left( \frac{1}{\sqrt{3}} \right)^4 = \frac{2}{9}. \end{align*} \] Now we can see that \(I(f) - G_2(f) \neq 0\) when the degree is 4. Hence, we can know tht \(m = 4\). Once, we know that \(w_1 = w_2 = 1\), and \(m = 4\), we can find the constant \(C\). Let \(f(x) = x^4\), we have \[ I(f) - G_2(f) = \frac25 - \frac29 = \frac{8}{45} = C f^{(4)}(\xi), \] for some \(\xi \in (-1, 1)\). Since \(f^{(4)}(x) = 24\) when \(f(x) = x^4\), we have \(f^{(4)}(\xi) = 24\). Therefore, we have \(C = \dfrac{8}{45 \cdot 24} = \dfrac{1}{135}\).

Solution (c). The degree of precision of \(G_2(f)\) is 3 because the approximation is exact when \(f(x) = 1, x, x^2, x^3\) but not exact when \(f(x) = x^4\).