(MSU) Solving Difference Equations (UNO)

The Gamma Function (XULA) Laguerre Polynomials (LSU) Solving Laplace (MSU) Solving Difference Equations (UNO) Stephen SaraMargaret Mladenka Kimbely Ward Tri Ngo

The Gamma Function - Research Forthcoming

The Laguerre Polynomial el.jpg The Laguerre equation is given by: ty + (1 t)y + ny = 0 where y(0) = 1 and n is non-negative integer

The Laguerre Polynomial 1 The Laguerre Polynomial of order n l n (t) = n k=0 n!( 1) k t k (k!) 2 (n k)! 2 For n m, e t l n (t)l m (t)dt = 0 0

The Laguerre Polynomials The first few polynomials are: 1 l 0 = 1 2 l 1 = t + 1 3 l 2 = 1 2 (t2 4t + 2) 4 l 3 = 1 6 ( t3 + 9t 2 18t + 6)

The Laguerre Polynomials on a Graph C:/Latek/Laguerre_Graph.jpg

Differential Operators and Relationships 1 E = td 2 + D 2 E l n = nl n 1 3 E O = 2tD 2 + (2 2t)D 1 4 E O l n = (2n + 1)l n 5 E + = td 2 + (1 2t)D + (t 1) 6 E + l n = (n + 1)l n+1 7 LIE BRACKET [E 0, E + ] = E 0 E + E + E 0 = 2E 0

The Laplace Transform 1 L n = L{l n (t)} = (s 1)n s n+1 2 L{l n (at)} = (s a)n s n+1

Laguerre Polynomial Properties t 1 0 l n(x)dx = l n (t) l n+1 (t) t 2 0 l n(x)l m (t x)dx = l m+n (t) l m+n+1 (t) 3 l n+1 (t) = 1 n+1 [(2n + 1 t)l n(t) nl n 1 (t)]

Solving Laplace Using DE s The Laplace Transform of a function, F (t), is f (s) = Therefore, Theorem If f (s) = L{F (t)}, then 1 sf (s) F (0) = L{F (t)} 2 f (s) = L{ tf (t)} 0 e at F (t)dt.

Theorems Theorem If X (t) At α (α > 1) as t 0, then x(s) AΓ(α + 1) s α+1 as s Theorem If X (t) Bt β (β > 1) as t, x(s) BΓ(β + 1) s β+1 as s 0

Table of Laplace f (s) F (t) x X sx X (0) X s 2 x sx (0) X (0) X x tx sx x tx s 2 x 2sx + X (0) tx x t 2 X sx + 2x t 2 X s 2 x + 4sx + 2x t 2 X Table: Basic Laplace

Laplace Transform Example Theorem The Laplace Transform of the zeroth-order Bessel Function is the following: 1 L{J 0 (t)} = s 2 + 1 and the Laplace Transform of the first-order Bessel Function is: L{J 1 (t)} = s 2 + 1 s s 2 + 1

Proving the Theorem Proof of Theorem. Given Equation X (t) = J 0 (a t) Differential Equation 4tX + 4X + a 2 X = 0 Laplace Transform x + 4s a2 x = 0 4s 2 Integrating Factor I = e a2 4s s Solve Diff. Equation Use Theorem 2 to find c. Plug in c = 1 a 2 ce 4s x(s) = s x(s) 1/s as s L{J 0 (a t)} = e a 4s 2 s.

Table of Inverse Laplace f (s) F (t) x X x tx x t 2 X sx X (0) X sx tx X sx t 2 X + 2tX s 2 x sx (0) X (0) X s 2 x + X (0) tx 2X s 2 x t 2 X + 4tX + 2X Table: Inverse Laplace

Inverse Laplace Transform Example Theorem The Bessel Differential Equation of the n th order can be defined as t 2 X + tx + (t 2 n 2 )X = 0 where the solutions of this differential equation are the n th order Bessel Functions.

Proving the Theorem Proof of Theorem. Given Equation x(s) = ( s 2 +1 s) n s 2 +1 Diff. Equation (s 2 + 1)x + 3sx + (1 n 2 )x = 0 Inverse Laplace t 2 X + tx + (t 2 n 2 )X = 0 Bessel s Equation X (t) = J n (t)

Solving Difference Equations a n+2 + ba n+1 + ca n = f (n) y(t) = a n, f (t) = f (n), for n t < n + 1 y(t + 2) + by(t + 1) + cy(t) = f (t) L{y(t + 1)} = e s Y (s) a 0e s (1 e s ) s L{y(t + 2)} = e 2s Y (s) es (1 e s )(a 0 e s + a 1 ) s L{f (t)} = 1 e s e sk f (k) s k=0

Solving Difference Equations - 2 Theorem 1. 2. 3. a n = Aα n + Bβ n a n = (An + B)α n r 1 = α + iβ, r 2 = α iβ a n = u n (A cos(nθ) + B sin(nθ)) with u = α 2 + β 2, θ = sin 1 β u

Solving Difference Equations - 3 a n+2 a n+1 a n = 0, a 0 = 0, a 1 = 1 Y (s) = es (1 e s ) s Y (s) = es (1 e s ) s(α β) = 1 ( e s s(α β) a n = a n = 1 5 1 (e s α)(e s β) ( 1 e s α 1 e s β ) 1 1 αe s 1 1 βe s 1 α β (αn β n ) [( 1 + ) n ( 5 2 ) 1 ) n ] 5 2

Solving Difference Equations - 4 Theorem Let a n,p be a fixed particular solution to the second order linear recursion relation a n+2 + ba n+1 + ca n = f (n). Then any other solution has the form a n = a n,h + a n,p, for some homogeneous solution a n,h.

Solving Difference Equations - 5 a n+2 + ba n+1 + ca n = f (n) (e 2s + be s + c)y (s) = p(e s ) sq(e s ) Y (s) = 1 s Y (s) = 1 s e sk H(k, p) = k=0 k=1 p(e s ) (1 rk e s ) p e sq c i (1 r k e s ) p 1 (1 re s ) p r n p 1 (p 1)! (n + k) = H(n, p) = L 1 1 e s { s(1 re s ) p }

Solving Difference Equations - 6 a n+2 5a n+1 + 6a n = n2 n, a 0 = 0, a 1 = 1 L{f (t)} = 1 e s s = 1 e s s = 1 e s s L{f (t)} = 1 e s s e sk k2 k k=0 ( ) e sk (k + 1)2 k e sk 2 k k=0 k=0 k=0 ( ) e sk H(k, 2) (2e s ) k k=0 ( ) 1 (1 2e s ) 2 1 1 2e s

Solving Difference Equations - 7 Y (s)(e 2s 5e s + 6) = 1 e s s 1 Y (s) = 1 e s s ( 1 (1 2e s ) ) 2 1 2e ( 3 1 3e s 5 2(1 2e s ) ) 1 2(1 2e s ) 3 s + es a n = 3 n+1 5 2 2n 1 H(n, 3){r = 2} 2 a n = 3 n+1 (n 2 + 3n + 12).2 n 2

Solving Difference Equations - 8 Theorem a n+2 + ba n+1 + ca n = g(n)r n, x 2 + bx + c = 0 deg(g) = k 1. 2. 3. a n,p = g (n)r n, deg(g ) = k a n,p = g (n)r n, deg(g ) = k + 1 a n,p = g (n)r n, deg(g ) = k + 2

Solving Difference Equations - 9 a n+2 + ba n+1 + ca n = c 1 sin(kn) + c 2 cos(kn) a n,p = c 3 sin(kn) + c 4 cos(kn)

Solving Difference Equations - Conclusion a n+2 4a n+1 + 4a n = (n 2 + 1)2 n + n3 n + 2 sin(4n) + 3 cos(n) a n = g(n)2 n +h(n)3 n +c 1 sin(4n)+c 2 cos(4n)+c 3 sin(n)+c 4 cos(n) with deg(g) = 4 and deg(h) = 1.

Contact Email: SaraMargaret Mladenka - smlade1@tigers.lsu.edu Tri Ngo - tmngo1@uno.edu Kimberly Ward - kaw205@msstate.edu Stephen - swilli13@xula.edu Website: http://www.math.lsu.edu/ rfrith1/math4999.html