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Truncated Exponential Series Inequality

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Truncated Exponential Series Inequality

Check out John Cook's explanation of truncated exponential series inequality.

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

T_n(x) = 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}

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 Tn(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) > Tn(x) if n is odd and exp(x) < Tn(x) if n is even.

Define fn(x) = exp(x) – Tn(x). If x > 0 then fn(x) > 0.

We want to show that if x < 0 then fn(x) > 0 for odd n and fn(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) ≠ 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) ≠ 0 for x < 0. So is  f2(x) always positive or always negative? This time we have

f2(x) = exp(x) – 1 – xx2/2

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

For general n, we assume fn 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 fn(x) is never zero for negative x. We then look at the dominant term –xn to argue that fn is positive or negative depending on whether n is odd or even.

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

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