/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 From $$P_{L}(\cos \theta)=\lef... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 6

From $$P_{L}(\cos \theta)=\left.\frac{1}{L !} \frac{\partial L}{\partial t^{L}}\left(1-2 t \cos \theta+t^{2}\right)^{-1 / 2}\right|_{t=0}$$ show that $$P_{L}(1)=1, \quad P_{L}(-1)=(-1)^{L} .$$

Short Answer

Expert verified
Thus, \( P_{L}(1) = 1 \) and \( P_{L}(-1) = (-1)^L \).

Step by step solution

01

Understanding the Legendre Polynomial

The Legendre polynomial, denoted as \( P_L(\cos \theta) \), is defined as a specific derivative expression. We are given: \[ P_{L}(\cos \theta) = \frac{1}{L!} \frac{\partial^L}{\partial t^L}(1 - 2t \cos \theta + t^2)^{-1/2} \bigg|_{t=0} \]Our task is to evaluate this at \( \theta = 0 \) and \( \theta = \pi \), which correspond to \( \cos \theta = 1 \) and \( \cos \theta = -1 \).
02

Evaluate the Expression for \( \cos \theta = 1 \)

Substituting \( \cos \theta = 1 \) in the expression:\[ P_{L}(1) = \frac{1}{L!} \frac{\partial^L}{\partial t^L}(1 - 2t + t^2)^{-1/2} \bigg|_{t=0} \]Simplifying inside the parentheses:\[ (1 - 2t + t^2) = (1 - t)^2 \]So the expression becomes:\[ (1 - t^2)^{-1/2} = (1 - t)^{-1} \]Thus,\[ P_{L}(1) = \frac{1}{L!} \frac{\partial^L}{\partial t^L}(1 - t)^{-1} \bigg|_{t=0} \]Taking the derivatives and evaluating them at \( t = 0 \), we find:\[ P_{L}(1) = 1 \] because the infinite series expansion leads to \( 1 \) for the first term.
03

Evaluate the Expression for \( \cos \theta = -1 \)

Similarly, substitute \( \cos \theta = -1 \) into the expression:\[ P_{L}(-1) = \frac{1}{L!} \frac{\partial^L}{\partial t^L}(1 + 2t + t^2)^{-1/2} \bigg|_{t=0} \]This simplifies to:\[ (1 + 2t + t^2) = (1 + t)^2 \]The derivative expression becomes:\[ (1 + t)^{-1} \]Thus,\[ P_{L}(-1) = \frac{1}{L!} \frac{\partial^L}{\partial t^L}(1 + t)^{-1} \bigg|_{t=0} \]Taking derivatives here leads to alternating signs which follow a pattern equivalent to \((-1)^L\). Hence,\[ P_L(-1) = (-1)^L \].

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 91影视!

Key Concepts

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

Partial Derivatives
Partial derivatives are a central concept when working with functions of multiple variables. They represent the rate at which a function changes as one particular variable changes, while others are held constant. In the context of Legendre Polynomials, the expression \[ \frac{\partial^L}{\partial t^L}(1 - 2t \cos \theta + t^2)^{-1/2} \bigg|_{t=0} \]is a higher-order partial derivative. This implies taking a derivative \( L \) times concerning the parameter \( t \), and then evaluating this derivative at \( t=0 \).
  • Each derivative taken simplifies the expression, stripping away the variables until you're left with a form that can be directly evaluated at \( t=0 \).
  • Interestingly, since we are dealing with higher-order derivatives, the expression often forms patterns or series, such as the binomial series, which heavily influences the result.
The key here is understanding how rapidly and meticulously one calculation follows another in sequences, revealing the remarkable patterns in special functions.
Polynomial Evaluation
Polynomial evaluation is about substituting values into a polynomial and calculating the result. When working with Legendre Polynomials, the task involves substituting specific values for \( \cos \theta \), which effectively evaluates these polynomials at particular points.
  • For example, substituting \( \cos \theta = 1 \) and \( \cos \theta = -1 \) simplifies our expression to more manageable polynomials, which translate to \( (1-t)^2 \) and \( (1+t)^2 \), respectively.
  • The idea is to simplify complex expressions into forms where derivatives can be easily computed, such as turning a higher degree expansion involving multiple variables into a single degree polynomial like \( (1-t)^{-1} \).
This simplification is crucial because it reduces computational complexity, allowing us to handle complicated polynomial behavior with ease and accuracy.
Special Functions
Special functions, like Legendre Polynomials, have unique properties and occasional symmetries that make them especially useful in various fields, such as physics and engineering. They are part of a broader class of functions called orthogonal polynomials, which have properties that make them extremely beneficial.
  • In particular, Legendre Polynomials are solutions to Legendre's differential equation, and they are orthogonal over the interval \([-1, 1]\) with a weight function of \(1\).
  • Their orthogonality helps in expanding functions as infinite series involving these polynomials, leading to more simplified integration or differentiation tasks in applied mathematics and physics.
  • When evaluated at the limits (\( \cos \theta = 1 \) and \( \cos \theta = -1 \)), they showcase integer properties like \( P_{L}(1) = 1 \) and \( P_{L}(-1) = (-1)^L \), revealing much about symmetry and alternating patterns inherent in these functions.
These attributes make special functions indispensable tools across scientific domains for modeling and simplifying complex systems.

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

Expand \(x^{8}\) as a Legendre series. Determine the Legendre coefficients from Eq. \((12.50)\), $$ a_{m}=\frac{2 m+1}{2} \int_{-1}^{1} x^{8} P_{m}(x) d x . $$ Check your values against AMS-55, Table 22.9. This illustrates the expansion of a simple function. Actually, if \(f(x)\) is expressed as a power series, the technique of Exercise \(12.2 .14\) is both faster and more accurate.

An atomic electron with angular momentum \(L\) and magnetic quantum number \(M\) has a wavefunction $$ \psi(r, \theta, \varphi)=f(r) Y_{L}^{M}(\theta, \varphi) $$ Show that the sum of the electron densities in a given complete shell is spherically symmetric, that is, \(\sum_{M=-L}^{L} \psi^{*}(r, \theta, \varphi) \psi(r, \theta, \varphi)\) is independent of \(\theta\) and \(\varphi\).

Neutrons (mass 1) are being scatt茅red by a nucleus of mass \(A(A>1)\). In the center of the mass system the scattering is isotropic. Then, in the lab system the average of the cosine of the angle of deflection of the neutron is $$ \langle\cos \psi\rangle=\frac{1}{2} \int_{0}^{\pi} \frac{A \cos \theta+1}{\left(A^{2}+2 A \cos \theta+1\right)^{1 / 2}} \sin \theta d \theta . $$ Show by expansion of the denominator that \((\cos \psi)=2 / 3 A\).

(a) Given \(f(x)=2.0,|x|<0.5 ; f(x)=0,0.5<|x|<1.0 .\) Expand \(f(x)\) in a Legendre series and calculate the coefficients \(a_{n}\) through \(a_{80}\) (analytically). (b) Evaluate \(\sum_{n=0}^{80} a_{n} P_{n}(x)\) for \(x=0.400(0.005) 0.600 .\) Plot your results. Note. This illustrates the Gibbs phenomenon of Section \(14.5\) and the danger of trying to calculate with a series expansion in the vicinity of a discontinuity.

Verify that the Legendre functions of the second kind, \(Q_{n}(x)\), satisfy the same recurrence relations as \(P_{n}(x)\), both for \(|x|<1\) and for \(|x|>1:\) $$ \begin{aligned} (2 n+1) x Q_{n}(x) &=(n+1) Q_{n+1}(x)+n Q_{n-1}(x), \\ (2 n+1) Q_{n}(x) &=Q_{n+1}^{\prime}(x)-Q_{n-1}^{\prime}(x) \end{aligned} $$
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Wave Optics

Read Explanation

Electricity and Magnetism

Read Explanation

Solid State Physics

Read Explanation

Force

Read Explanation

Modern Physics

Read Explanation
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.