Define fn :R → R by fn(x) = (nC0 + nC2 x + nC4 x2 + ... ) / (nC1 + nC3 x + nC5 x2 + ... ), where nCm is the binomial coefficient. Find a formula for fn+1(x) in terms of fn(x) and x, and determine limn→∞fn(x) for all real x.
Solution
It is almost obvious that fn+1 = (fn + x)/(fn + 1) (*). So if fn(x) tends to a limit k(x), then k(x) = ( k(x) + x) /( k(x) + 1), and hence k(x) = √x.
Obviously, fn(0) = 1/n, so k(0) = 0. We notice also that fn(x) = (√x) N/D, where N = (1 + √x)n + (1 - √x)n and D = (1 + √x)n - (1 - √x)n.
Suppose 0 < x ≤ 1. Then put y = (1 - √x)/(1 + √x). We have 0 ≤ y < 1 and N/D = (1 + yn)/(1 - yn) which tends to 1, so in this case also k(x) = √x.
Similarly if x > 1, put y = (√x - 1)/(1 + √x) and we again get k(x) = √x.
It is clear from (*), that for x < 0, fn(x) does not tend to a limit.
© John Scholes
jscholes@kalva.demon.co.uk
25 Jan 2002