Show that \(\int_{-1}^{1} x^{m} P_{l}(x) d x=0\) if \(m

Short Answer

Expert verified
The integral evaluates to zero due to the orthogonality of Legendre polynomials when \( m < l \).

Step by step solution

01

- Understand Rodrigues' formula

Rodrigues' formula for Legendre polynomials is given by \[ P_l(x) = \frac{1}{2^l l!} \frac{d^l}{dx^l} (x^2 - 1)^l. \] This will be used for further steps.
02

- Initial integral setup

The integral to evaluate is \[ I = \int_{-1}^{1} x^m P_l(x) \, dx. \]
03

- Apply integration by parts

Use integration by parts where \[ u = x^m \text{ and } dv = P_l(x) \, dx. \] Then, \[ du = m x^{m-1} \, dx \text{ and } v = \frac{1}{2^l l!} \int P_l(x) \, dx. \]
04

- Repeatedly differentiate and integrate

Notice that differentiating powers of \( x \) decreases the degree, while integrating the polynomial retains the highest degree term. After repeatedly applying integration by parts, a term with \( P_{l-1}(x) \) will appear.
05

- Examine boundary conditions

Since \( P_l(x) \) is a polynomial of degree \( l \), and the Legendre polynomial satisfies orthogonality, any polynomial of lower degree, here \( x^m \) with \( m < l \), will result in the integral equating to zero.
06

- Conclusion

Therefore, \[ \int_{-1}^{1} x^m P_l(x) \, dx = 0 \] if \( m < l \). The orthogonality condition of Legendre polynomials confirms this.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rodrigues' formula
Rodrigues' formula is fundamental for understanding Legendre polynomials. It provides a way to generate these polynomials systematically. The formula states: \[ P_l(x) = \frac{1}{2^l l!} \frac{d^l}{dx^l} (x^2 - 1)^l \]Here,
  • \(P_l(x)\) is the Legendre polynomial of degree \(l\)
  • \(2^l\) indicates that the exponent corresponds to the polynomial’s degree
  • \(l!\) is the factorial of \(l\)
  • \(\frac{d^l}{dx^l}\) represents the \(l-\text{th}\) derivative with respect to \(x\)

This formula shows how to derive \(P_l(x)\) using the derivatives of \((x^2 - 1)^l\). Each Legendre polynomial can thus be expressed in terms of these derivatives, providing a powerful tool for further calculations.
By inserting into integrals or differential equations, Rodrigues' formula streamlines the process and lays the groundwork for more advanced mathematical operations.
Integration by Parts
Integration by parts is a key technique to solve integrals involving products of functions. The rule is generally articulated as: \[ \int u \, dv = uv - \int v \, du\]where
  • \(u\) and its derivative \(du\) are parts of one function
  • \(dv\) and its antiderivative \(v\) are parts of another function

Use this method to tackle the problem where \[ I = \int_{-1}^{1} x^m P_l(x) \, dx \]Assign \(u = x^m\) and \(dv = P_l(x) \, dx\). Differentiating and integrating these parts repeatedly, recalling that differentiating powers of \(x\) decreases the degree, leads to simpler functions.
This gradual breakdown generates new integrals easier to solve or reveals boundary terms that evaluate to zero due to specific limits.
This rule is crucial to simplify and evaluate the integral provided in the exercise.
Orthogonality
Orthogonality in the context of polynomials means that the integral of the product of two different polynomials over a specific interval is zero. For Legendre polynomials, this property is mathematically expressed as: \[ \int_{-1}^{1} P_l(x) P_m(x) \, dx = 0, \text{ for } l eq m \]In our exercise, it proves that \( \int_{-1}^{1} x^m P_l(x) \, dx = 0 \) if \( m < l \).
Since \(P_l(x)\) is a polynomial of degree \(l\), any polynomial of lesser degree, like \(x^m\) where \( m < l \), when multiplied and integrated over the same interval with \(P_l(x)\), results in zero.
This orthogonality is a major simplification when dealing with polynomial integrals and is widely used in solving differential equations, series expansions, and more.
These integral properties arise from the symmetry and definition of Legendre polynomials, making them invaluable in mathematical analysis.

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Most popular questions from this chapter

Solve the following differential equations by series and also by an elementary method and verify that your solutions agree. Note that the goal of these problems is not to get the answer (that's easy by computer or by hand) but to become familiar with the method of series solutions which we will be using later. Check your results by computer. $$y^{\prime}=3 x^{2} y$$

Express each of the following polynomials as linear combinations of Legendre polynomials. Hint: Start with the highest power of \(x\) and work down in finding the correct combination. \(5-2 x\)

Show in two ways that \(\int_{-1}^{1} P_{2 n+1}(x) d x=0\).

By power series, solve the Hermite differential equation $$y^{\prime \prime}-2 x y^{\prime}+2 p y=0$$ You should find an \(a_{0}\) series and an \(a_{1}\) series as for the Legendre equation in Section 2 Show that the \(a_{0}\) series terminates when \(p\) is an even integer, and the \(a_{1}\) series terminates when \(p\) is an odd integer. Thus for each integer \(n\), the differential equation (22.14) has one polynomial solution of degree \(n\). These polynomials with \(a_{0}\) or \(a_{1}\) chosen so that the highest order term is \((2 x)^{n}\) are the Hermite polynomials. Find \(H_{0}(x), H_{1}(x),\) and \(H_{2}(x) .\) Observe that you have solved an eigenvalue problem (see end of Section 2 ), namely to find values of \(p\) for which the given differential equation has polynomial solutions, and then to find the corresponding solutions (eigenfunctions).

Prove the least squares approximation property of Legendre polynomials [see (9.5) and (9.6)] as follows. Let \(f(x)\) be the given function to be approximated. Let the functions \(p_{l}(x)\) be the normalized Legendre polynomials, that is, $$p_{l}(x)=\sqrt{\frac{2 l+1}{2}} P_{l}(x) \quad \text { so that } \quad \int_{-1}^{1}\left[p_{l}(x)\right]^{2} d x=1$$ Show that the Legendre series for \(f(x)\) as far as the \(p_{2}(x)\) term is $$f(x)=c_{0} p_{0}(x)+c_{1} p_{1}(x)+c_{2} p_{2}(x) \quad \text { with } \quad c_{l}=\int_{-1}^{1} f(x) p_{l}(x) d x$$ Write the quadratic polynomial satisfying the least squares condition as \(b_{0} p_{0}(x)+\) \(b_{1} p_{1}(x)+b_{2} p_{2}(x)\) (by Problem 5.14 any quadratic polynomial can be written in this form). The problem is to find \(b_{0}, b_{1}, b_{2}\) so that $$I=\int_{-1}^{1}\left[f(x)-\left(b_{0} p_{0}(x)+b_{1} p_{1}(x)+b_{2} p_{2}(x)\right)\right]^{2} d x$$ is a minimum. Square the bracket and write \(I\) as a sum of integrals of the individual terms. Show that some of the integrals are zero by orthogonality, some are 1 because the \(p_{t}\) 's are normalized, and others are equal to the coefficients \(c_{l}\). Add and subtract \(c_{0}^{2}+c_{1}^{2}+c_{2}^{2}\) and show that $$I=\int_{-1}^{1}\left[f^{2}(x)+\left(b_{0}-c_{0}\right)^{2}+\left(b_{1}-c_{1}\right)^{2}+\left(b_{2}-c_{2}\right)^{2}-c_{0}^{2}-c_{1}^{2}-c_{2}^{2}\right] d x$$ Now determine the values of the \(b\) 's to make \(I\) as small as possible. (Hint: The smallest value the square of a real number can have is zero.) Generalize the proof to polynomials of degree \(n\).

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