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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

By computer, find 30 zeros of \(J_{0}\) and note that the spacing between consecutive zeros is tending to \(\pi\).

Show that any polynomial of degree \(n\) can be written as a linear combination of Legendre polynomials with \(l \leq n\)

Evaluate \(\int_{-1}^{1} P_{0}(x) P_{2}(x) d x\) to show that these functions are orthogonal on (-1,1).

Solve the differential equations by the Frobenius method; observe that you get only one solution. (Note, also, that the two values of \(s\) are equal or differ by an integer, and in the latter case the larger \(s\) gives the one solution.) Show that the conditions of Fuchs's theorem are satisfied. Knowing that the second solution is \(\ln x\) times the solution you have, plus another Frobenius series, find the second solution. $$4 x^{2}(x+1) y^{\prime \prime}-4 x^{2} y^{\prime}+(3 x+1) y=0$$

Consider each of the following problems as illustrations showing that, in a power series solution, we must be cautious about using the general recursion relation between the coefficients for the first few terms of the series. Solve \(y^{\prime \prime}+y^{\prime} / x^{2}=0\) by power series to find the relation $$a_{n+1}=-\frac{n(n-1)}{n+1} a_{n}.$$ If, without thinking carefully, we test the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) for convergence by the ratio test, we find $$\lim _{n \rightarrow \infty} \frac{\left|a_{n+1} x^{n+1}\right|}{\left|a_{n} x^{n}\right|}=\infty\quad (Show this.)$$ Thus we might conclude that the series diverges and that there is no power series solution of this equation. Show why this is wrong, and that the power series solution is \(y=\) const.

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free