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Show that any polynomial of degree n can be written as a linear combination of Legendre polynomials with ln

Short Answer

Expert verified
A polynomial of degree n can be expressed as a linear combination of Legendre polynomials of degrees l with ln using orthogonal properties and integration.

Step by step solution

01

Understand Legendre Polynomials

Legendre polynomials are a set of orthogonal polynomials which arise in solving certain types of differential equations. They are denoted as Pl(x) where l is a non-negative integer.
02

Define the Polynomial

Consider a polynomial P(x) of degree n, which can be written as P(x)=a0+a1x+a2x2+...+anxn
03

Express Polynomial as a Linear Combination

To show that P(x) can be expressed as a linear combination of Legendre polynomials, assume P(x)=c0P0(x)+c1P1(x)+c2P2(x)+...+cnPn(x) where ci are constants and Pi(x) are Legendre polynomials of degree i.
04

Use Orthogonality

Legendre polynomials are orthogonal, meaning 11Pi(x)Pj(x)dx=0forieqj. This orthogonality helps in finding the coefficients ci.
05

Compute the Coefficients

Multiply both sides of the polynomial equation by Pk(x) for a fixed k and integrate over the interval [1,1]. 11P(x)Pk(x)dx=11(i=0nciPi(x))Pk(x)dx Due to orthogonality, all terms where ieqk vanish, leaving 11P(x)Pk(x)dx=ck11(Pk(x))2dx
06

Solve for the Constants

Solve for ck by isolating it: ck=11P(x)Pk(x)dx11(Pk(x))2dx
07

Conclusion

Using the calculated coefficients ck, substitute them back into the linear combination, showing that the polynomial of degree n can be written as a sum of Legendre polynomials with degrees up to n.

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Key Concepts

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

Orthogonal Polynomials
Orthogonal polynomials, like Legendre polynomials, have unique properties that make them useful in many fields of study such as physics and engineering.
These polynomials are 'orthogonal' over a specified interval with respect to a weight function. This means that their inner product is zero when integrating the product of two different polynomials over that interval.
For Legendre polynomials, this interval is [1,1] and the weight function is 1.
Orthogonality simplifies many calculations, especially when dealing with polynomial approximations.
Because of orthogonality, it becomes easier to decompose complex functions into simpler, orthogonal components.
Polynomial Decomposition
Polynomial decomposition involves breaking down a complex polynomial into simpler components, often a linear combination of simpler polynomials.
This is particularly useful when analyzing the behavior of the polynomial or solving equations.
In the context of Legendre polynomials, any polynomial of degree n can be expressed as P(x)=c0P0(x)+c1P1(x)+c2P2(x)+...+cnPn(x)
Here, ci are coefficients, and Pi(x) are Legendre polynomials of degree i.
This decomposition makes it easier to handle complex polynomials by leveraging the orthogonal characteristics of Legendre polynomials.
Coefficients Calculation
Calculating the coefficients in the polynomial decomposition is an essential step.
For this, we use the orthogonality property of Legendre polynomials.
To find a specific coefficient, say ck, we multiply the entire polynomial by Pk(x) and integrate over the interval [1,1].
Due to orthogonality, the integral simplifies significantly, resulting in the formula: ck=11P(x)Pk(x)dx11(Pk(x))2dx
This formula isolates ck, allowing us to determine the coefficients needed for the polynomial decomposition.
Orthogonality in Polynomials
Orthogonality in polynomials provides a powerful tool for polynomial analysis and decomposition.
When polynomials are orthogonal, as with Legendre polynomials, it simplifies many integral calculations.
The orthogonality condition, 11Pi(x)Pj(x)dx=0for iej,
means that the integral of the product of two different orthogonal polynomials over the interval [1,1] is zero.
This property is extremely useful in calculating coefficients and simplifies the process of expressing complex polynomials as linear combinations of orthogonal polynomials.

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Most popular questions from this chapter

Verify that the differential equation x4y+y=0 is not Fuchsian; that it has the two independent solutions xsin(1/x) and xcos(1/x); and that these solutions are not expandable in Frobenius series.

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. xy=xy+y

Solve the following differential equations by the method of Frobenius (generalized power series). Remember that the point of doing these problems is to learn about the method (which we will use later), not just to find a solution. You may recognize some series [as we did in (11.6)] or you can check your series by expanding a computer answer. 2xy+y+2y=0

Show that R=lx(1x2)D and L=lx+(1x2)D, where D=d/dx, are raising and lowering operators for Legendre polynomials [compare Hermite functions, (22.1) to (22.11) and Bessel functions, Problems 22.29 and 22.30]. More precisely, show that RPl1(x)=lPl(x) and LPl(x)=lPl1(x). Hint: Use equations (5.8d) and (5.8f). Note that, unlike the raising and lowering operators for Hermite functions, here R and L depend on l as well as x, so you must be careful about indices. The L operator operates on Pl, but the R operator as given operates on Pl1 to produce lPl. [If you prefer, you could replace l by l+1 to rewrite R as (l+1)x(1x2)D then it operates on Pl to produce (l+1)Pl+1. ] Assuming that all Pl(1)=1, solve LP0(x)=0 to find P0(x)=1, and then use raising operators to find P1(x) and P2(x).

Solve the following differential equations by the method of Frobenius (generalized power series). Remember that the point of doing these problems is to learn about the method (which we will use later), not just to find a solution. You may recognize some series [as we did in (11.6)] or you can check your series by expanding a computer answer. x2y6y=0

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