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 *f*_{1} and its derivative are both zero at 0. Now suppose *f*_{1} 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 *f*_{1} is 0. Since the derivative of *f*_{1} is also zero at 0, there must be some point *c* with *b* < *c* < 0 where the second derivative of *f*_{1} is 0, again by Rolle’s theorem. But the second derivative of *f*_{1} is exp(*x*) which is never 0. So our assumption *f*_{1}(*a*) = 0 leads to a contradiction.

Now *f*_{1}(0) = 0 and *f*_{1}(*x*) ≠ 0 for *x* < 0. So *f*_{1}(*x*) must be always positive or always negative. Which is it? For negative *x*, exp(*x*) is bounded and so

*f*_{1}(*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 *f*_{2}(*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 *f*_{2} 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 *f*_{2} is exp(*x*) which is never zero.

As before the contradiction shows *f*_{2}(*x*) ≠ 0 for *x* < 0. So is *f*_{2}(*x*) always positive or always negative? This time we have

*f*_{2}(*x*) = exp(*x*) − 1 − *x* − *x*^{2}/2

which is eventually dominated by the −*x*^{2} 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 −*x*^{n} 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.

I’m wondering if this follows from Taylor’s Theorem with the Lagrange form of the remainder. As I recall, that form is based on the Mean Value Theorem as well.

It’s always amazing to me how many theorems in analysis have as a fundamental step “exp(x) is never 0”. It seems to be a pretty fundamental fact.

Thank you for this post.

I usually thought that why would I care to post such elementary proofs of facts but today I wanted to know whether this is true or not and googled to find your post.

Also, you gave pretty interesting proof. Another way can be to use induction with the hypothesis that

f_(2m-1)(x) > 0 for x <0 and f_(2m)(x) < 0 for x < 0.

I find it pretty cool, which I realized from your post is that derivative of f_(2m)(x) is f_(2m-1)(x) and so on. Kinda cool!