Frobenious

# Frobenious - Frobenius method example for Sept 20 lecture...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Frobenius method example for Sept. 20 lecture Preliminary remarks; using the Gamma function. The Gamma function often appears in the solution to recursion equations for series solutions to o.d.e. The Gamma function Γ( x ) is defined for all reals numbers x except x = 0, x =- 1, x =- 2, etc., where it has vertical asymptotes. For x > 0 it is defined by the formula Γ( x ) = Z ∞ t x- 1 e- tdt. For x = n , n a positive integer, Γ( n ) = ( n- 1)!. Integration by parts shows that Γ( x +1) = x Γ( x ) , x > , and Γ( x ) is extended to x < 0 (except 0 and the negative integers) by imposing this identity everywhere. The usefulness of the Gamma function for recursions is due to the following cal- culation. Let k be a positive integer and let p be a real number. Then Γ( p +1) = p Γ( p ) = p ( p- 1)Γ( p- 1) = p ( p- 1)( p- 2)Γ( p- 2) = ··· = p ( p- 1)( p- 2) ··· ( p- k )Γ( p- k ) . This is valid as long as none of p,p- 1 ,...,p- k are non-negative integers or zero. Dividing both sides of this identity by Γ( p- k ), leads to the identity: p ( p- 1)( p- 2) ··· ( p- k ) = Γ( p +1) Γ( p- k ) . (1) 2. We solve x 2 y ” + ( x/ 2) y + (1 / 2) x 2 y = 0 in a series solution about the regular singular point x = 0. We look for two linearly independent solutions. The method of Frobenius says to look for solutions of the form | x | r ∞ X a n x n , where a = 1 and r is to be determined from the indicial equation. In doing the calculations, it is convenient to assume x > 0, so that | x | r = x r , and to write y = ∞ X a n x r + n . To do this, calculate x 2 y ” and xy ” by term by term differentiation and plug into the equation. The result is 0 = ∞ X ( r + n )( r + n- 1) a n x r + n + ∞ X (1 / 2)( r + n ) a n x r + n + ∞ X ( a n / 2) x r + n +2 . (2) 1 To combine the series, they should all be expressed in powers of r + n . Only the last series fails to conform, so we fix it by changing the index of summation to k = n + 2, and then relabeling the new index k using n again: ∞ X n =0 ( a n / 2) x r + n +2 = ∞ X k =2 ( a k- 2 / 2) x r + k...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

Frobenious - Frobenius method example for Sept 20 lecture...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online