A general continuing fraction recursion algorithm for square roots
The continuing real fraction
J + 1/(J + 1/(J + 1/(J + 1/ ...
= J + [(J^2 + 4)^.5]/2
is a special case of a recursion function yielding that limit. That general function is
Xsub(n+1) = (Xsub(n) + C)^(-1) + C
Setting Xo = 0 and C = J, we see (where sub(-1) is not an initial value but a designation for the constant prior to application of the function):
Convergent ; Our function; Continued fraction0 ; Xsub(-1) = J; J
1 ; Xsub(1) = (J^-1) + J; J + J^-1
2 ; Xsub(2) = (J + J^-1)^-1 + J; as above
Of course, we needn't set Xo = 0. In fact, the curious thing is that this recursion function arrives at the same limit no matter what real initial value is chosen (other than Xo = -C, which must be excluded).
That is, (lim n-->inf)Xsub(n+1) = (lim n-->inf)Ysub(n+1)
when Xsub(1) = (Xo + C)^-1 + C and Ysub(1) = (Yo + C)^-1 + C. It is the constant C that determines the limit, which is the limit of the continuing fraction
1 + 1/C...
That is, beginning with any real but -C for Xo and any real but -C for Yo, we obtain the limit above because we find that
(lim n-->inf)(Xsub(n) - Ysub(n)) = 0,
where (Xsub(n) - Ysub(n)) alternates sign by n.
A bit of perfunctory algebra, which I omit, establishes these facts.
So, this algorithm yields an infinity of approaches to any square root. That is, Xsub(n) =/= Ysub(n) for finite n.
An example: (lim n-->inf)X(sub n) = (2 + 8^.5)/2 = 1 + 2^.5
For Xo = 1 and C = 2, some recursive (calculator) values are:
3
2.333...
2.428571429
2.411764706
2.414634146
For Xo = 1/2 and C = 2
2.5 2.4 2.416...6...
2.413793103
2.414285714
For Xo = -31 and C = 2
-29.0
1.965517241
2.50877193
2.398601399
2.416909621
For Xo = 31 and C = 2
33.0
2.03...03...
2.492537313
2.401197605
2.416458853
For Xo = 1/31 and C = 2 2.032258065
2.492063492
2.401273885
2.416445623
2.413830955
For Xo = -1/31 and C = 2
1.967741935
2.508196721
2.39869281
2.416893733
Note the pattern of alternately too high--too low.
No comments:
Post a Comment