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Q29E

Expert-verifiedFound in: Page 450

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**The equation**

**(1-x ^{2})y"-2xy'+n(n+1)y=0**

**where n is an unspecified parameter is called Legendre’s equation. This equation appears in applications of differential equations to engineering systems in spherical coordinates.**

**(a) Find a power series expansion about x=0 for a solution to Legendre’s equation. **

**(b) Show that for a non negative integer there exists an n ^{th} degree polynomial that is a solution to Legendre’s equation. These polynomials upto a constant multiples are called Legendre polynomials.**

**(c) Determine the first three Legendre polynomials (upto a constant multiple).**

(a) The power series expansion is y(x)=a_{0} [1+Σ_{k=1}^{∞ } (-1)^{k} [n(n-2)(n-4)...(n-2k+2)(n+1)(n+3)...(n+2k-1)]/(2k)! x^{2k}] +a_{1} [x+Σ_{k=1}^{∞ } (-1)^{k} [(n-1)(n-3)...(n-2k+1)(n+2)(n+4)...(n+2k)]/(2k+1)! x^{2k+1}]. .

(b) We showed that for a nonnegative integer there exists a Legendre polynomial.

(c) The first three Legendre polynomials are P_{0}(x)=1, P_{1}(x)=x, P_{2}(x)=1-3/2x^{2}.

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a recurrence relation for the coefficients.

**A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable. **

It is generally given by the formula,

y(x)=Σ_{n=0}^{∞ }a_{n}x

Given,

(1-x^{2})y"-2xy'+n(n+1)y=0

Let,

y(x)=Σ_{k=0}^{∞ }a_{k}x^{k}

Taking derivative of the above equation,

y'(x)=Σ_{k=1}^{∞ }ka_{k}x^{k-1}

y"(x)=Σ_{k=2}^{∞ }k(k-1)a_{k}x^{k-2}

Replace in the equation.

(1-x^{2})Σ_{k=2}^{∞ }k(k-1)a_{k}x^{k-2} -2x Σ_{k=1}^{∞ }ka_{k}x^{k-1} +n(n+1)Σ_{k=0}^{∞ }a_{k}x^{k} =0

Σ_{k=2}^{∞ }k(k-1)a_{k}x^{k-2} -Σ_{k=2}^{∞ }k(k-1)a_{k}x^{k}-2 Σ_{k=1}^{∞ }ka_{k}x^{k} +n(n+1)Σ_{k=0}^{∞ }a_{k}x^{k} =0

The first sum substitute is,

Σ_{k=0}^{∞} (m+2)(m+1)a_{m+2}x^{m} -Σ_{k=2}^{∞ }k(k-1)a_{k}x^{k} -2Σ_{k=1}^{∞ }ka_{k}x^{k}+n(n+1)Σ_{k=0}^{∞ }a_{k}x^{k} =0

(2)(1)a_{2}+(3)(2)a_{3}+Σ_{k=2}^{∞} (m+2)(m+1)a_{m+2}x^{m} - Σ_{k=2}^{∞ }k(k-1)a_{k}x^{k} - (2)(1) a_{1}x -2Σ_{k=1}^{∞ }ka_{k}x^{k} + n(n+1) (a_{0}+a_{1}x) +n(n+1)Σ_{k=0}^{∞ }a_{k}x^{k} =0

2a_{2}+n(n+1)a_{0} + (6a_{3}-2a_{1}+n(n+1)a_{1}) x+Σ_{k=2}^{∞} [(k+2)(k+1)a_{k+2}+(-k^{2}+k-2k+n(n+1))a_{k}] x^{k }=0

2a_{2}+n(n+1)a_{0} + (6a_{3}+(n-1)(n+2)a_{1}) x+Σ_{k=2}^{∞} [(k+2)(k+1)a_{k+2}+(n(n+1)-k(k+1)) a_{k}] x^{k }=0

Hence the first sum substitute is 2a_{2}+n(n+1)a_{0} + (6a_{3}+(n-1)(n+2)a_{1}) x+Σ_{k=2}^{∞} [(k+2)(k+1)a_{k+2}+(n(n+1)-k(k+1)) a_{k}] x^{k }=0.

Coefficients must equal zero.

2a_{2}+n(n+1)a_{0}=0

a_{2}= -n(n+1)/2 a_{0}

6a_{3}+(n-1)(n+2)=0

a_{3}= -(n-1)(n+2)/6 a_{1}

(k+2)(k+1)a_{k+2}+(n(n+1)-k(k+1)) a_{k}=0

a_{k+2}= -[n(n+1)-k(k+1)]/(k+2)(k+1) a_{k}

= -[(n+k+1)(n-k)]/(k+2)(k+1) a_{k}

a_{4}=a_{2+2}

= -[(n+3)(n-2)]/(4)(3) a_{2}

=(-1)^{2} [(n-2)n(n+1)(n+3)]/4! a_{0}

a_{5}=a_{3+2}

= -[(n+4)(n-3)]/(5)(4) a_{3}

=(-1)^{2} [(n-3)(n-1)(n+2)(n+3)]/5! a_{0}

Now you can find general formula for coefficients.

a_{2k}=(-1)^{k} [n (n-2)(n-4)...(n-2k+2) (n+1)(n+3)...(n+2k-1)]/2k! k≥1

a_{2k+1 }= (-1)^{k} [(n-1)(n-3)...(n-2k+1) (n+2)(n+4)...(n+2k)]/(2k+1)! k≥1

Hence the general solution is,

y(x)=a_{0} [1+Σ_{k=1}^{∞} (-1)^{k} [n (n-2)(n-4)...(n-2k+2) (n+1)(n+3)...(n+2k-1)]/2k! ] x^{2k} +a_{1} [x+Σ_{k=1}^{∞} (-1)^{k} [(n-1)(n-3)...(n-2k+1) (n+2)(n+4)...(n+2k)]/(2k+1)! ] x^{2k+1} k≥1

Let’s find the radius of convergence for this series.

|a_{k+2}/a_{k}|<1

|(n+k+1) (n-k)/(k+2)(k+1)| <1

This is satisfied for choosing k=n. Therefore the series cut in specific integers n and n+1 produce polynomials called Legendre polynomials.

Let’s find Legendre polynomial for n=0.

P_{0}(x)=1

For n=1, k can be 1 or zero.

P_{1}(x)=x

For n=2, k can be.

P_{2}(x)=1+(-1) (2+1)/[(2)(1)!] x^{2}

=1-3/2 x^{2}

The Legendre polynomials are P_{0}(x)=1, P_{1}(x)=x, and P_{2}(x)=1-3/2 x^{2 }.

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