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I have a question about Spivak's Calculus, chapter 10 problem 19. I feel I must be missing something very simple, but I don't know what it is.

Here is the question:

  1. Prove that if $f^{n}(g(a))$ and $g^{n}(a)$ both exist, then $(f \circ g)^{n}(a)$ exists. A little experimentation should convince you that it is unwise to seek a formula for $(f \circ g)^{n}(a)$. In order to prove that $(f \circ g)^{n}(a)$ exists you will therefore have to devise a reasonable assertion about $(f\circ g)^{n}(a)$ which can be proved by induction.

Here is my conjecture.

$\textbf{Conjecture}$. If $g^{n}(a)$ and $f^{n}(g(a))$ both exist, then $(f\circ g)^{n}(a)$ exists and is a sum of products of the form:

$$c[g{'}(a)]^{m_{1}}[g^{2}(a)]^{m_{2}} \dots [g^{n}(a)]^{m_{n}}f^{k}(g(a)) $$

where $c$ is a constant, $m_{1}, \dots , m_{n}$ are natural numbers, and $k$ is a natural number such that $k\leq n$.

The base case for $n=1$ is straightforward. If $g{'}(a)$ exists and $f{'}(g(a))$ exists, then the Chain Rule tells us that $(f\circ g)^{'}(a)$ exists and is equal to $g{'}(a)f{'}(g(a))$. Thus, the conjecture holds for $n=1$ if we let $c=m_{1}=k=1$.

Next we assume:

$\textbf{Inductive Hypothesis}$ If $g^{n}(a)$ and $f^{n}(g(a))$ exist, then $(f\circ g)^{n}(a)$ exists and is a sum of products of the form:

$$c[g{'}(a)]^{m_{1}}[g^{2}(a)]^{m_{2}} \dots [g^{n}(a)]^{m_{n}}f^{k}(g(a)) $$

We now want to show: If $g^{n+1}(a)$ exists and $f^{n+1}(g(a))$ exists, then $(f\circ g)^{n+1}(a)$ exists and is a sum of products of the form (where $k\leq n+1$):

$$c[g{'}(a)]^{m_{1}}[g^{2}(a)]^{m_{2}} \dots [g^{n}(a)]^{m_{n}}[g^{n+1}(a)]^{m_{n+1}}f^{k}(g(a)) $$

I am confused about the answers that I have found in the solution book and on stack exchange here: Spivak Chapter 10, Exercise 19 solution verification. Prove that if $f^{(n)}(g(a))$ and $g^{(n)}(a)$ both exist, then $(f\circ g)^{(n)}(a)$ exists..

For the inductive step, the author shows that if $f^{n+1}(g(a))$ exists and $g^{n+1}(a)$ exist, then it follows that for all x in some interval around a, $f^{n}(g(x))$ and $g^{n}(x)$ both exist. They then conclude from the Inductive Hypothesis that, in this interval, $(f\circ g)^{n}$ exists and is equal to a sum of products of the form:

$$c[g{'}(x)]^{m_{1}}[g^{2}(x)]^{m_{2}} \dots [g^{n}(x)]^{m_{n}}f^{k}(g(x)) $$

Here the author is not talking about $(f\circ g)^{n}$ at a single point a. Instead they are talking about $(f \circ g)^{n}(x)$ at all points in an interval around $a$. This the part that I don't understand. $\textbf{Why do we need to establish that $(f\circ g)^{n}(x)$ exists for all x in some interval around $a$?}$ Why can't we just say: If $g^{n+1}(a)$ exists, then $g^{n}(a)$ exists. By the definition of $g^{n+1}$, we have:

$$g^{n+1}(a) = \lim_{h\to 0}\frac{g^{n}(a+h) - g^{n}(a)}{h}$$

So $g^{n}(a)$ must exist. Similarly, we know that $f^{n}(g(a))$ exists since:

$$f^{n+1}(g(a)) = \lim_{h\to 0}\frac{f^{n}(g(a) + h) - f^{n}(g(a))}{h}$$

So by the Inductive Hypothesis, it follows that $(f\circ g)^{n}(a)$ exists and is a sum of terms of products of the form:

$$c[g{'}(a)]^{m_{1}}[g^{2}(a)]^{m_{2}} \dots [g^{n}(a)]^{m_{n}}f^{k}(g(a)) $$

It is then easy to use the product, chain, and sum rules for the derivative to show that $(f\circ g)^{n+1}(a) is a sum of products that all have the required form.

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    $\begingroup$ Were either of these answers helpful? If so, you might want to accept one of them. $\endgroup$
    – Ben
    Commented Nov 6, 2022 at 13:18

2 Answers 2

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Let's recall some definitions first:

Definition $1$: Interior point of a set $S\subset \mathbb R$:

An element $s\in S$ is said to be an interior point of $S$ if there exists an $r>0$ such that the open interval $(s-r,s+r)\subset S$. (i.e., there is an open interval (containing $s$) which is contained in $S$).

Definition $2$: Interior of a set $S\subset \mathbb R$:

The set of all interior points of $S$ is said to be interior of $S$. Let's denote this by $S^o$.

Let $D$ be a subset of $\mathbb R$ with non -empty interior.

Now, let's define derivative of a function $h:D \to \mathbb R$ at $a\in D^o$:-

Definition $3$: We say that $h$ is differentiable at $a$ iff the the limit $\lim_{x\to a}\frac{h(x)-h(a)}{x-a}$ exists.

Remark: Note that the definition requires $a$ to be an interior point of domain of $h$.

๐–๐ก๐ฒ ๐๐จ ๐ฐ๐ž ๐ง๐ž๐ž๐ ๐ญ๐จ ๐ž๐ฌ๐ญ๐š๐›๐ฅ๐ข๐ฌ๐ก ๐ญ๐ก๐š๐ญ $(๐‘“\circ ๐‘”)^{(๐‘›)}(๐‘ฅ)$ ๐ž๐ฑ๐ข๐ฌ๐ญ๐ฌ ๐Ÿ๐จ๐ซ ๐š๐ฅ๐ฅ ๐ฑ ๐ข๐ง ๐ฌ๐จ๐ฆ๐ž ๐ข๐ง๐ญ๐ž๐ซ๐ฏ๐š๐ฅ ๐š๐ซ๐จ๐ฎ๐ง๐ $๐‘Ž$?

This is to ensure that $(f\circ g)^{(n+1)}(a)$ makes sense by the definition above. By the remark after the definition, $a$ should be an interior point of domain of $(f\circ g)^{(n)}$, which by definition of an interior point means that there should be an $r>0$ such that $(a-r,a+r)\subset $ domain of $(f\circ g)^{(n)}$.

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Suppose we have three functions $f$, $g$, and $h$ such that at a particular point $a$ we have $$f(a) = cg(a) + kh(a),$$ where $c$ and $k$ are constants.

Suppose also that the functions $g$ and $h$ are each differentiable at $a$.

Is this enough to conclude $$f'(a) = cg'(a) + kh'(a)?$$

What if we also know that $$f(x) = cg(x) + kh(x),$$ for all $x$ in some interval containing $a$?

Keeping these ideas in mind, let's return to your concluding statement:

So by the Inductive Hypothesis, it follows that $(f\circ g)^{n}(a)$ exists and is a sum of terms of products of the form: $$c[g{'}(a)]^{m_{1}}[g^{2}(a)]^{m_{2}} \dots [g^{n}(a)]^{m_{n}}f^{k}(g(a))$$ It is then easy to use the product, chain, and sum rules for the derivative to show that $(f\circ g)^{n+1}(a)$ is a sum of products that all have the required form.

Can you see the error here, and the utility of knowing that

$$(f\circ g)^{n}(x) = c[g{'}(x)]^{m_{1}}[g^{2}(x)]^{m_{2}} \dots [g^{n}(x)]^{m_{n}}f^{k}(g(x)),$$ for all $x$ in some interval around (and including) $a$?

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