Truncated exponential series inequality

Define T_n to be the Taylor series for exp(x) truncated after n terms:

How does this function compare to its limit, exp(x)? We might want to know because it's often useful to have polynomial upper or lower bounds on exp(x).

For x > 0 it's clear that exp(x) is larger than T_n(x) since the discarded terms in the power series for exp(x) are all positive.

The case of x < 0 is more interesting. There exp(x) > T_n(x) if n is odd and exp(x) < T_n(x) if n is even.

Define f_n(x) = exp(x) - T_n(x). If x > 0 then f_n(x) > 0.

We want to show that if x < 0 then f_n(x) > 0 for odd n and f_n(x) < 0 for even n.

For n = 1, note that f₁ and its derivative are both zero at 0. Now suppose f₁ is zero at some point a < 0. Then by Rolle's theorem, there is some point b with a < b < 0 where the derivative of f₁ is 0. Since the derivative of f₁ is also zero at 0, there must be some point c with b < c < 0 where the second derivative of f₁ is 0, again by Rolle's theorem. But the second derivative of f₁ is exp(x) which is never 0. So our assumption f₁(a) = 0 leads to a contradiction.

Now f₁(0) = 0 and f₁(x) a 0 for x < 0. So f₁(x) must be always positive or always negative. Which is it? For negative x, exp(x) is bounded and so

f₁(x) = exp(x) - 1 - x

is eventually dominated by the -x term, which is positive since x is negative.

The proof for n = 2 is similar. If f₂(x) is zero at some point a < 0, then we can use Rolle's theorem to find a point b < 0 where the derivative of f₂ is zero, and a point c < 0 where the second derivative is zero, and a point d < 0 where the third derivative is zero. But the third derivative of f₂ is exp(x) which is never zero.

As before the contradiction shows f₂(x) a 0 for x < 0. So is f₂(x) always positive or always negative? This time we have

f₂(x) = exp(x) - 1 - x - x²/2

which is eventually dominated by the -x² term, which is negative.

For general n, we assume f_n is zero for some point x < 0 and apply Rolle's theorem n+1 times to reach the contradiction that exp(x) is zero somewhere. This tells us that f_n(x) is never zero for negative x. We then look at the dominant term -xⁿ to argue that f_n is positive or negative depending on whether n is odd or even.

Another way to show the sign of f_n(x) for negative x would be to apply the alternating series theorem to x = -1.