4.7 Higher order derivatives

Let $S$ be a subset of $R$ (or $C$ ), and Let $f:S\to C$ be a function of a real (or complex) variable. We say that $f$ is continuously differentiable on $S^0$ if $f$ is differentiable at each point $x$ of $S^0$ and the function $f^{\text{'}}$ is continuous on $S^0.$ We say that $f\in C^1\left(S\right)$ if $f$ is continuous on $S$ and continuously differentiable on $S^0.$ We say that $f$ is 2-times continuously differentiable on $S^0$ if the first derivative $f^{\text{'}}$ is itself continuously differentiable on $S^0.$ And, inductively, we say that $f$ is k-times continuously differentiable on $S^0$ if the $k-1$ st derivative of $f$ is itself continuously differentiable on $S^0.$ We write $f^{\left(k\right)}$ for the $k$ th derivative of $f,$ and we write $f\in C^k\left(S\right)$ if $f$ is continuous on $S$ and is $k$ times continuously differentiable on $S^0.$ Of course, if $f\in C^k\left(S\right),$ then all the derivatives $f^{\left(j\right)}$ , for $j\le k,$ exist nd are continuous on $S^0.$ (Why?)

For completeness, we define $f^{\left(0\right)}$ to be $f$ itself, and we say that $f\in C^{\infty }\left(S\right)$ if $f$ is continuous on $S$ and has infinitely many continuous derivatives on $S^0;$ i.e., all of its derivatives exist and are continuous on $S^0.$

As in [link] , we say that $f$ is real-analytic (or complex-analytic ) on $S$ if it is expandable in a Taylor series around each point $c\in S^0$

REMARK Keep in mind that the definition above, as applied to functions whose domain $S$ is a nontrivial subset of $C,$ has to do with functions of a complex variable that are continuously differentiable on the set $S^0.$ We have seen that this is quite different from a function having continuous partial derivatives on $S^0.$ We will return to partial derivatives at the end of this chapter.

Let $S$ be an open subset of $R$ (or $C$ ).

1. Suppose $WS$ is a subset of $R.$ Then, for each $k\ge 1,$ there exists a function in $C^k\left(S\right)$ that is not in $C^{k+1}\left(S\right).$ That is, $C^{k+1}\left(S\right)$ is a proper subset of $C^k\left(S\right).$
2. If $f$ is real-analytic (or complex-analytic) on $S,$ then $f\in C^{\infty }\left(S\right).$
3. There exists a function in $C^{\infty }\left(R\right)$ that is not real-analytic on $R.$ That is, the set of real-analytic functions on $R$ is a proper subset of the set $C^{\infty }\left(R\right).$

REMARK Suppose $S$ is an open subset of $C.$ It is a famous result from the Theory of Complex Variables that if $f$ is in $C^1\left(S\right),$ then $f$ is necessarily complex analytic on $S.$ We will prove this amazing result in [link] . Part (3) of the theorem shows that the situation is quite different for real-valued functions of a real variable.

For part (1), see the exercise below. Part (2) is immediate from part (c) of [link] . Before finishing the proof of part (3), we present the following lemma:

Let $f$ be the function defined on all of $R$ as follows.

where $p\left(x\right)$ is a fixed polynomial function and $n$ is a fixed nonnegative integer. Then $f$ is continuous at each point $x$ of $R.$

The assertion of the lemma is clear if $x\ne 0.$ To see that $f$ is continuous at 0, it will suffice to prove that

(Why?) But, for $x>0,$ we know from part (b) of [link] that $e^{1/x}>1/(x^{n+1}(n+1)!),$ implying that $e^{-1/x}<x^{n+1}(n+1)!.$ Hence, for $x>0,$

and this tends to 0 as $x$ approaches 0 from the right, as desired.

Returning to the proof of [link] , we verify part (3) by observing that if $f$ is as in the preceding lemma then $f$ is actually differentiable, and its derivative $f^{\text{'}}$ is a function of the same sort. (Why?) It follows that any such function belongs to $C^{\infty }\left(R\right).$ On the other hand, a nontrivial such $f$ cannot be expandable in a Taylor series around 0because of the Identity Theorem. (Take $x_k=-1/k.$ ) This completes the proof.

Formula for the coefficients of a taylor series function

Let $f$ be expandable in a Taylor series around a point $c:$

Then for each $n,$ $a_n=f^{\left(n\right)}\left(c\right)/n!.$

Because each derivative of a Taylor series function is again a Taylor series function, and because the value of a Taylor series function at the point $c$ is equal to its constant term $a_0,$ we have that $a_1=f^{\text{'}}\left(c\right).$ Computing the derivative of the derivative, we see that $2a_2={f^{\text{'}}}^{\text{'}}\left(c\right)=f^{\left(2\right)}\left(c\right).$ Continuing this, i.e., arguing by induction, we find that $n!a_n=f^{\left(n\right)}\left(c\right),$ which proves the theorem.