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.
Best Answer
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.