1992, A-4

Alternative solution:
Let g(x) = 1/(1 + x^2).  We may compute the derivatives as before, and again
the problem reduces to showing that all derivatives of f(x)-g(x) = h(x)
vanish at 0.

By continuity, h^(0) vanishes at 0.  We now proceed by induction using the
nth mean value theorem, which states that h(x) = h(0) + h'(0) + ... +
h^(n-1)(0)/(n-1)! + h^(n)(\theta)/n!, where 0 <= \theta <= x.  By induction,
the derivatives up to the (n-1)-th vanish at 0, and if we take x = 1, 1/2,
... we get h^(n)(\theta_n) = 0, where 0 <= \theta_n <= 1/n.  Hence \{\theta_n\}
tends to 0.  But h is infinitely differentiable, so h^n is infinitely
differentiable and hence continuous.  It follows that h^(n)(0) = 0.

Yet another solution:
Consider only n divided by l.c.m/(1,2,\ldots,k).
f^(k)(0)   = \lim_{n\rightarrow\infty}
   ( \sum_{i=0}^k {k\choose i}(-1)^{i+1} f(i/n)  ) / (1/n)^k
           = \lim_{n\rightarrow\infty}
   ( \sum_{i=0}^k {k\choose i}(-1)^{i+1} g(i/n)  ) / (1/n)^k
=g^{k}(0)= \cos(\pi k/2) k!

Or replace n by n*l.c.m.(1,2,\ldots,k) in the above and allow n to be any
positive integer.



Source: sol.cgi/competition/tests/math/putnam/putnam.1992