A CLASS ASSIGNED PROBLEM FOR PHYSICS 555ASPRING 2008. CSUF

BY NASSER ABBASI

1 Problem

Show that the recurence formula

2(k− q) Cq = − q(q-+-2l+-1)(k-+-l)Cq−1

(1)

can be written as

q( 2 )q (k− 1)! (2l+ 1)! Cq = (− 1) k-+l (k−-q−-1)!q!(q-+2l+-1)!C0

(2)

2 Solution

Proof by induction on $q$ . For $q = 1$ , equation (1) becomes

2(k− 1) C1 = −-------------C0 (2l+ 2)(k+ l)

and equation (2) becomes

Hence it is true for $q = 1$ . Now assume it is true for $q = n$ , in otherwords, assume that

-----2(k−-n)----- Cn = − n(n + 2l+ 1)(k + l)Cn−1

(3)

implies

( 2 )n (k− 1)! (2l+ 1)! Cn = (− 1)n ---- -----------------------C0 k +l (k− n− 1)!n!(n +2l+ 1)!

(4)

Now for the induction step. we need to show that it is true for $n+ 1$ , i.e. given (4) is true, we need to show that, by replacing $n$ by $n + 1$ in the above, that

C = − -------2-(k-−-(n-+-1))-------C n+1 (n+ 1)((n+ 1)+ 2l+ 1)(k+ l) n

(5)

implies

We start with (5), and replace the $C n$ term with what we assumed to be true from (4), hence (5) can be rewritten as

Cnfrom(4) ◜[------(-----)n-----◞◟------------------]◝ Cn+1 = −--------2(k−-(n+-1))------- (− 1)n --2- --(k−-1)!- --(2l+-1)!--C0 (n + 1)((n +1 )+ 2l+1 )(k + l) k + l (k− n − 1)! n!(n + 2l+ 1)!

Simplify the above leads to

( ) n+1 --2- n+1 -(k−-1)!-----(2l+-1)!--- Cn+1 = (− 1) k+ l (k − n− 2)!n!(n+ 2l+ 2)!C0

Which is (6). Therefore, the relationship is true for any $n$ . QED