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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 the real numbers of degree 2n for which p(x) greater than or equal to 0 for all x, then p(x) + p'(x) + p''(x) + ... + p(2n)(x) greater than or equal to 0 for all x.  This may be proved similarly to the above.

Source: Traditional

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