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Chapter 5: Problem 13

Use the zeros of the Legendre polynomial \(P_{2}(x)\) to obtain a two-point quadrature formula $$\int_{-1}^{1} f(x) d x \approx A_{1} f\left(x_{1}\right)+A_{2} f\left(x_{2}\right)$$

Short Answer

Expert verified
The two-point quadrature formula based on the zeros of the Legendre polynomial \(P_2(x)\) is: $$\int_{-1}^1 f(x) dx \approx f\left(-\frac{1}{\sqrt{3}}\right) + f\left(\frac{1}{\sqrt{3}}\right)$$

Step by step solution

01

Find the zeros of the Legendre polynomial \(P_2(x)\)

To find the zeros of the Legendre polynomial \(P_2(x)\), we need to solve the equation \(P_2(x) = 0\). We first need to determine the polynomial \(P_2(x)\). The second-degree Legendre polynomial is given by: $$P_2(x) = \frac{1}{2}(3x^2 - 1)$$ Setting \(P_2(x) = 0\), we have: $$\frac{1}{2}(3x^2 - 1) = 0$$ Let's solve for \(x\).
02

Solve for the zeros \(x_1\) and \(x_2\)

To solve for \(x\), we can first multiply by 2 to get rid of the fraction: $$3x^2 - 1 = 0$$ Next, we can add 1 to both sides and then divide by 3: $$x^2 = \frac{1}{3}$$ Taking the square root of both sides gives the two zeros: $$x_1 = -\frac{1}{\sqrt{3}}$$ $$x_2 = \frac{1}{\sqrt{3}}$$
03

Compute the weights \(A_{1}\) and \(A_{2}\)

To compute the weights \(A_{1}\) and \(A_{2}\), we need to solve the following system of linear equations, arising from requiring the quadrature formula to integrate the basis polynomials, \(1\) and \(x\), exactly: $$A_{1} + A_{2} = \int_{-1}^1 1 dx$$ $$A_{1}x_{1} + A_{2}x_{2} = \int_{-1}^1 x dx$$ By evaluating these integrals, we get: $$A_{1} + A_{2} = 2$$ $$A_{1}x_{1} + A_{2}x_{2} = 0$$ We can now solve this system of equations to find the weights \(A_{1}\) and \(A_{2}\).
04

Solve for the weights \(A_{1}\) and \(A_{2}\)

Using the previously calculated values of \(x_1\) and \(x_2\), and the system of equations from Step 3, we can solve for the weights \(A_1\) and \(A_2\). Substitute the values of \(x_1\) and \(x_2\) into the second equation: $$A_{1}\left(-\frac{1}{\sqrt{3}}\right) + A_{2}\left(\frac{1}{\sqrt{3}}\right) = 0$$ From the first equation, we can express \(A_1 = 2 - A_2\). Substitute this into the second equation and solve for \(A_2\): $$\left(2 - A_2\right)\left(-\frac{1}{\sqrt{3}}\right) + A_2\left(\frac{1}{\sqrt{3}}\right) = 0$$ Simplified, we get: $$A_{2} = 1$$ Now, we can find \(A_1\): $$A_{1} = 2 - A_{2} = 2 - 1 = 1$$
05

Write down the quadrature formula

Now that we have the zeros and weights, we can write down the two-point quadrature formula: $$\int_{-1}^1 f(x) dx \approx A_{1}f\left(x_{1}\right) + A_{2}f\left(x_{2}\right) = f\left(-\frac{1}{\sqrt{3}}\right) + f\left(\frac{1}{\sqrt{3}}\right)$$

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Most popular questions from this chapter

Let \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) be an orthonormal basis for an inner product space \(V\) and let $$\mathbf{u}=\mathbf{u}_{1}+2 \mathbf{u}_{2}+2 \mathbf{u}_{3} \quad \text { and } \quad \mathbf{v}=\mathbf{u}_{1}+7 \mathbf{u}_{3}$$ Determine the value of each of the following: (a) \(\langle\mathbf{u}, \mathbf{v}\rangle\) (b) \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\) (c) The angle \(\theta\) between \(\mathbf{u}\) and \(\mathbf{v}\)

Let \(\mathbf{u}\) be a unit vector in \(\mathbb{R}^{n}\) and let \(H=I-2 \mathbf{u u}^{T}\) Show that \(H\) is both orthogonal and symmetric and hence is its own inverse.

Let \(A\) be an \(m \times 3\) matrix. Let \(Q R\) be the \(Q R\) factorization obtained when the classical GramSchmidt process is applied to the column vectors of \(A,\) and let \(\tilde{Q} \tilde{R}\) be the factorization obtained when the modified Gram-Schmidt process is used. Show that if all computations were carried out using exact arithmetic, then we would have $$\tilde{Q}=Q \quad \text { and } \quad \tilde{R}=R$$ and show that, when the computations are done in finite-precision arithmetic, \(\tilde{r}_{23}\) will not necessarily be equal to \(r_{23}\) and consequently \(\tilde{r}_{33}\) and \(\tilde{\mathbf{q}}_{3}\) will not necessarily be the same as \(r_{33}\) and \(\mathbf{q}_{3}\)

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