Solving Recurrence Relations Using Differential Operators - Sections
Solving Recurrence Relations Using Differential Operators - Sections
Solving Recurrence Relations Using Differential Operators - Sections
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We begin by multipling the relation by xn<br />
n! and then sum it up from n = 0 to<br />
n = 1. Thus,<br />
1X<br />
n=0<br />
an+1x n<br />
n!<br />
=<br />
1X (n + 1)anxn : (28)<br />
n!<br />
n=0<br />
We distribute the right hand side of the equation to get<br />
1X<br />
n=0<br />
an+1x n<br />
n!<br />
=<br />
1X<br />
n=0<br />
nanx n<br />
n! +<br />
1X anxn : (29)<br />
n!<br />
By resetting our indices we can use equations (9), (10), and (11) to get<br />
This is a separable di¤erential equation:<br />
Integrating, we obtain<br />
<strong>Solving</strong> for h(x) we get<br />
n=0<br />
h 0 (x) = xh 0 (x) + h(x): (30)<br />
h 0 (x)<br />
h(x)<br />
<strong>Using</strong> the initial condition of h(0) = 1, we …nd<br />
1<br />
= : (31)<br />
1 x<br />
ln jh(x)j = ln j1 xj + C: (32)<br />
h(x) = eC<br />
: (33)<br />
1 x<br />
h(x) = 1<br />
: (34)<br />
1 x<br />
We now use the McLaurin series expansion of the right hand side of<br />
h(x) =<br />
1X<br />
x n : (35)<br />
n=0<br />
We need to get h(x) in the correct form so we mulitply by n!<br />
h(x) =<br />
Now we can solve for our an to get<br />
1X<br />
n=0<br />
n!xn : (36)<br />
n!<br />
an = n!: (37)<br />
6
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