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*to argue that

^{n}*f*is positive or negative depending on whether

_{n}*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.

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