Using Taylor expansion for a smooth real function

Question

I've come across the following problem in Cracking Mathematics Subject Test, 4th Edition by Steve Leduc, from Princeton Review.

Let $f(x)$ be a function that has derivatives of all orders at every real $x$, such that $f(0)=0$. Assume the following about the falues of the derivatives of $f$ at $x=0$: $f^\prime(0)=1$; $f^{\prime\prime}(0)=2$; and $f^{(n)}(0)\le n!/2^{n-3}$ for every integer $n\le 3$. What is the smallest number $m$ that makes the statement $f(1)\le m$ true? (2, 3, 4, 6, 8)

The "solution" in the book first uses the Taylor expansion with infinitely many terms (without a remainder term) of $f$ around 0, plug in the inequalities and $x=1$, and conclude $m=4$.

To my best knowledge, just because $f$ can be differentiated infinitely many times doesn't mean it has a Taylor expansion. Why can the "solution" be justified?

Another not-so-directly-related question is whether $g$ is equal to a power series $\sum_i a_ix^i$ on the interval $(-R,R)$, if $R$ is the radius of convergence of the series and the series coincides with $g$ on some interval containing 0.

I would be grateful if you could help me solve these questions.

Answer 1

Assuming you mean the inequality for n $\ge$ 3, the Taylor's expansion T, whether it converges to f or not, satisfies the inequality:

$T \le 1 + 1 + 2^3\sum_{k = 3}^\infty \frac{1}{2^k}$. This certainly converges to 1 + 1 + 8(1-1/2-1/4) = 1 + 1 + 2 = 4, which is no doubt where they got the m = 4.

However, we don't know that this series converges to f, so I don't think we can conclude anything about f(1).

In answer to your second question, if g is equal to the series on some small interval (-a,a) around 0, it can certainly be extended to all of (-R,R) simply by setting it equal to the series. However, given a g which is previously defined and coincides with the series on (-a,a) there is no reason at all why it should match the series outside the interval (-a,a). For example, g might equal the series on (-a,a) and be zero everywhere else.

Even if you say that the pre-defined g is infinitely differentiable, it is perfectly possible for it not to coincide with the series outside of (-a,a).

However, if you say that g is analytic, then it will match the given series throughout the entire interval of convergence. We can see that by thinking of g as a complex analytic function. A fundamental fact about analytic functions is that if g is analytic at the point 0 (or the point w) then there is a Taylor's expansion for it around the point 0 (or w), and the expansion is good throughout a circle around 0 (w) up to its radius of convergence (that is to say, the closest point where g is no longer analytic).

Since the Taylor's expansion is unique, if g matches the given series on (-a,a) then it must be the Taylor's expansion for g, and it will match g throughout |z| < R.

You can now restrict g to the x-axis, and you have your result.

Now you may ask, what if g as a complex analytic function does not extend to the entire circle |z| < R. Well, then the radius of convergence of the series cannot be R.

As for the book possibly containing an error, many books do.

Answer 2

The short answer to your second question is no: the standard example is the bump function $$f(x) = \begin{cases}e^{-1/x^2}, &x>0\\0, &x\leq 0\end{cases}$$ which is $C^\infty$ but has every derivative zero at $x=0$; clearly it does not converge to its Taylor series $f(x) = 0$ on any neighborhood of $0$.

However, Taylor's theorem with remainder is valid for any sufficiently smooth function, regardless of whether or not it is analytic -- but if the remainder terms grow too quickly as you add more and more terms, the infinite Taylor series might not converge. Often when you have strong bounds on your derivatives, you can work directly with the former and avoid questions of convergence.