Solution to puzzle 102: Almost exponential
Show that 1 + x + x2/2! + x3/3! + ... + x2n/(2n)! is positive for all real values of x.
Let p(x) = 1 + x + x2/2! + x3/3! + ... + x2n/(2n)!
Clearly p(x) > 0 and increasing for large absolute values of x.
Since p is a polynomial, it is a continuous function, and so assumes a minimum value (or possibly more than one minimum) on the real numbers.
Let p reach a minimum at x = a. Then p'(a) = 0.
We have p'(x) = 1 + x + x2/2! + x3/3! + ... + x2n−1/(2n−1)!
Hence p(a) = p'(a) + a2n/(2n)! = a2n/(2n)!.
Since p'(0) = 1, p cannot reach a minimum at x = 0, and so p(a) = a2n/(2n)! > 0.
Hence p has a positive minimum.
Therefore, p(x) = 1 + x + x2/2! + x3/3! + ... + x2n/(2n)! is positive for all real values of x.
Remarks
The Maclaurin series for the exponential function, ex, is 1 + x + x2/2! + ... + xn/n! + ... .
More generally, if p(x) is a polynomial over 
of degree 2n for which p(x) 
0 for all x, then p(x) + p'(x) + p''(x) + ... + p(2n)(x) 0 for all x. This may be proved similarly to the above.
Source: Traditional