91Ó°ÊÓ

The Legendre polynomial of degree 2 is given by the formula \(P_{2}(x) = \frac{1}{2}(3x^{2} - 1)\). To find its zeros, we need to solve the equation \(P_{2}(x) = 0\): \begin{align*} \frac{1}{2}(3x^{2} - 1) &= 0 \\ 3x^{2} - 1 &= 0 \\ 3x^{2} &= 1 \\ x^{2} &= \frac{1}{3} \\ x &= \pm\sqrt{\frac{1}{3}} \end{align*} Thus, the zeros of the Legendre polynomial \(P_{2}(x)\) are \(x_1 = -\sqrt{\frac{1}{3}}\) and \(x_2 = \sqrt{\frac{1}{3}}\).

To find the weights \(A_1\) and \(A_2\), we use the property of the Legendre polynomial that the weights are equal since the polynomial is symmetric with respect to the \(y\)-axis: \[A_1 = A_2 = A\] Now we use the property that the quadrature formula is exact for polynomials of degree up to 3, which means it can accurately integrate a polynomial of degree 3. We will use a polynomial of degree 1 as an example, so let's consider a linear function \(f(x) = ax + b\). We have: \begin{align*} \int_{-1}^{1} (ax + b) d x &= A(ax_1 + b) + A(ax_2 + b) \\ \int_{-1}^{1} (ax + b) d x &= A(a(x_1 + x_2) + 2b) \end{align*} Now we know that the integral of a linear function over the interval \([-1,1]\) is given by: \[\int_{-1}^{1} (ax + b) d x = 2b\] Comparing the quadrature formula and the exact integral, we get: \[2b = A(a(x_1 + x_2) + 2b)\] Since this equation must hold for all \(a\) and \(b\), we can see that \(x_1 + x_2 = 0\), and \(A = 1\). This property is indeed satisfied: \[x_1 + x_2 = -\sqrt{\frac{1}{3}} + \sqrt{\frac{1}{3}} = 0\] Therefore, we have \(A_1 = A_2 = 1\).

Now that we have the zeros and weights, we can form the two-point quadrature formula: \[\int_{-1}^{1} f(x) d x \approx A_{1} f\left(x_{1}\right)+A_{2} f\left(x_{2}\right) = f\left(-\sqrt{\frac{1}{3}}\right) + f\left(\sqrt{\frac{1}{3}}\right)\] So, the two-point quadrature formula using the zeros of the Legendre polynomial \(P_{2}(x)\) is: \[\int_{-1}^{1} f(x) d x \approx f\left(-\sqrt{\frac{1}{3}}\right) + f\left(\sqrt{\frac{1}{3}}\right)\]