| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
7 - 5x[n]
let x[n+1] = --------- for 0 < x[0] < 1
8
The sequence x[n] converges to ... what? Is there a closed form for
the limit in the general case where 7, 5 and 8 are replaced by a,b,c?
I've obviously been away from school a lonnnnggg time.
John
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1958.1 | RTL::GILBERT | Wed Apr 05 1995 14:35 | 18 | ||
Let x[n+1] = a/c - x[n]*b/c.
If the series converges to x[inf], then
x[inf] = a/c - x[inf]*b/c, and so
x[inf] = a/(b + c).
Define delta by x[n] = a/(b + c) + delta. Then
x[n+1] = a/c - x[n]*b/c = a/(b+c) - delta*b/c
If |b/c| < 1, then the series converges, for any starting x[0].
We have the closed form solution:
x[n] = a/(b + c) + (x[0]-a/(b+c))*(-b/c)^n
BTW, for (a,b,c) = (7,5,8), x[inf] = 7/(5+8) = 7/13.
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| 1958.2 | Solution | FLOYD::YODER | MFY | Wed Apr 05 1995 14:46 | 8 |
If b+c=0, x[n] = x[0]+an/c
Otherwise
n
x[n] = a/(b+c) + (-b/c) (x[0] - a/(b+c))
The easy way to get to this is to note that if K=a/(b+c), then
cK + bK - a = 0; subtract this from cx[n+1] + bx[n] - a = 0 to get
a recurrence in x'[n] = x[n] - K.
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