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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. $$ 3 x^{2}+x-1 $$

Short Answer

Expert verified
The polynomial is expressed as the linear combination: \( 2P_2(x) + P_1(x) - P_0(x) \).

Step by step solution

01

Identify the Legendre Polynomials

The first few Legendre polynomials are: \( P_0(x) = 1 \)\( P_1(x) = x \)\( P_2(x) = \frac{1}{2}(3x^2 - 1) \)
02

Match the Highest Power

Observe that the highest power in the given polynomial is \(x^2\). Match it with the highest power in the Legendre polynomials:\( P_2(x) = \frac{1}{2}(3x^2 - 1) \).We need to match the \(3x^2\) term. Notice: \( 3x^2 = 3 \cdot \frac{1}{2}(3x^2 - 1) + \frac{3}{2} \)Thus, we can say: \( 3x^2 = 2P_2(x) + \frac{3}{2} \)
03

Simplify the Matching Coefficients

We obtained \( 3x^2 = 2P_2(x) + \frac{3}{2} \).Now subtract \( \frac{3}{2} \):\( 3x^2 - \frac{3}{2} = 2P_2(x) \)
04

Combine Terms to Form the Polynomial

Next, we include the remaining terms of the polynomial. The terms left are \(x - 1\). Rewriting:\[3x^2 + x - 1 = 2P_2(x) + x - 1 + \frac{3}{2} - \frac{3}{2}\]
05

Express Remaining Terms as Legendre Polynomials

We notice the polynomial can now be expressed with remaining Legendre polynomials:So, \( x \) remains as \( P_1(x) \) and \(- 1\) can be rewritten as:\( -1 = -1P_0(x) \)Thus the final expression becomes:\[ 3x^2 + x - 1 = 2P_2(x) + P_1(x) - P_0(x) \]

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Key Concepts

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

Legendre polynomials
Legendre polynomials are a sequence of orthogonal polynomials that solve certain types of differential equations. These polynomials are crucial in physics and engineering, especially in contexts involving spherical coordinates. Each Legendre polynomial is denoted by \( P_n(x) \), where \( n \) is the degree. For example:
  • \( P_0(x) = 1 \)
  • \( P_1(x) = x \)
  • \( P_2(x) = \frac{1}{2}(3x^2 - 1) \)
Notice how each polynomial includes terms of lower degrees? This hierarchy helps express other polynomials as a combination of Legendre polynomials.
Polynomial expression
A polynomial expression is a sum of terms, each consisting of a variable raised to a non-negative integer power and multiplied by a coefficient. For example, in the polynomial \( 3x^2 + x - 1 \), the terms are:
  • \( 3x^2 \) (second-degree term)
  • \( x \) (first-degree term)
  • \( -1 \) (constant term)
To express a given polynomial as a linear combination of Legendre polynomials, match each term in the target polynomial to forms in available Legendre polynomials.
Orthogonal polynomials
Orthogonal polynomials, like Legendre polynomials, are polynomials that remain orthogonal with respect to a specific weight function over a given interval. Orthogonality implies that the integral of the product of two different polynomials over a certain interval is zero: \[\[\begin{equation}otag \int_{-1}^{1} P_m (x) P_n (x) dx = 0 \end{equation}\]\] for \( m eq n \). Since Legendre polynomials are orthogonal, they simplify solving problems involving expansions of functions, making them quite useful in approximation techniques and simplifying complex problems into manageable forms.
Mathematical methods
Mathematical methods involving Legendre polynomials often include expressing functions as series, solving differential equations, or optimizing certain expressions. For instance, in our problem:

To express \( 3x^2 + x - 1 \) as a linear combination of Legendre polynomials:
  • Identify Legendre polynomials up to the necessary degree.
  • Match the highest power term to the corresponding Legendre polynomial.
  • Simplify and match lower power terms iteratively.
This approach breaks down the problem into easier, digestible steps, ensuring accuracy and understanding. Applying these methods gives the essential skills for tackling broader mathematical problems.

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Most popular questions from this chapter

Consider each of the following problems as illustrations showing that we must be cautious about using the general recursion relation between the coefficients in a power series solution 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 scries \(\sum_{n=0}^{\infty} a_{n} x^{n}\) for convergence by the ratio test, we find $$ \lim _{n \rightarrow \infty} \frac{\left|a_{n+1} x^{n+1}\right|}{\left|a_{n} x^{n}\right|}=\infty $$ (Show this.) Thus we might conclude that the series diverges and that there is no power series solution of this cquation. Show why this is wrong, and that the power series solution is \(y=\) const.

Solve the following differential equations by power series and also by an elementary method. Verify that the series solution is the power series expansion of your other solution. $$ y^{\prime}=x y+x $$

For each of the following equations, one solution \(u\) is given. Find the other solution by assuming a solution of the form \(y=u v\). \(\left(x^{2}+1\right) y^{n}-x y^{\prime}+y=0 ; u=x\)

Solve each of the following 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 general solution. (It is convenient to note that the value of \(s\) in the second Frobenius series is always the same as the second value of \(s\) which did not give a solution in the first part of the problem.) \(x y^{\prime \prime}+y^{\prime}=0\)

\(x^{2} y^{\prime \prime}+2 x y^{\prime}-6 y=0\)

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