The obvious question to ask about Taylor polynomials is ‘What are the first so-many terms in the Taylor polynomial of some function expanded at some point?’.

The most straightforward way to deal with this is just to do what is indicated by the formula: take however high order derivatives you need and plug in. However, very often this is not at all the most efficient.

Especially in a situation where we are interested in a composite function of the form $f(x^n)$ or $f(\hbox{polynomial in }x)$ with a ‘familiar’ function $f$, there are alternatives.

For example, looking at $f(x)=e^{x^3}$, if we start taking
derivatives to expand this at $0$, there will be a big mess pretty
fast. On the other hand, we might start with the ‘familiar’ expansion
for $e^x$
$$e^x=1+x+{x^2\over 2!}+{x^3\over 3!}+{e^c\over 4!}x^4$$
with some $c$ between $0$ and $x$, where our choice to cut it off
after that many terms was simply a whim. But then
replacing $x$ by $x^3$ gives
$$e^{x^3}=1+x^3+{x^6\over 2!}+{x^9\over 3!}+{e^c\over 4!}x^{12}$$
with some $c$ between $0$ and $x^3$. Yes, we need to keep track of $c$
in relation to the *new* $x$.

So we get a polynomial plus that funny term with the ‘c’ in it, for the remainder. Yes, this gives us a different-looking error term, but that's fine.

So we obtain, with relative ease, the expansion of degree
*eleven* of this function, which would have been horrible to
obtain by repeated differentiation and direct application of the
general formula. Why ‘eleven’?: well, the error term has the $x^{12}$
in it, which means that the polynomial itself stopped with a $x^{11}$
term. Why didn't we see that term? Well, evidently the coefficients of
$x^{11}$, and of $x^{10}$ (not to mention $x,x^2,x^4,x^5,x^7,x^8$!)
are *zero*.

As another example, let's get the degree-eight expansion of $\cos x^2$ at $0$. Of course, it makes sense to use $$\cos x=1-{x^2\over 2!}+{x^4\over 4!}+{-\sin c\over 5!}x^5$$ with $c$ between $0$ and $x$, where we note that $-\sin x$ is the fifth derivative of $\cos x$. Replacing $x$ by $x^2$, this becomes $$\cos x^2=1-{x^4\over 2!}+{x^8\over 4!}+{-\sin c\over 5!}x^{10}$$ where now we say that $c$ is between $0$ and $x^2$.

#### Exercises

- Use a shortcut to compute the Taylor expansion at $0$ of $\cos (x^5)$.
- Use a shortcut to compute the Taylor expansion at $0$ of $e^{(x^2+x)}$.
- Use a shortcut to compute the Taylor expansion at $0$ of $\log({ 1 \over 1-x })$.