# Spivak Calculus Chapter 10 problem 19

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.

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

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 $$h$$.

𝐖𝐡𝐲 𝐝𝐨 𝐰𝐞 𝐧𝐞𝐞𝐝 𝐭𝐨 𝐞𝐬𝐭𝐚𝐛𝐥𝐢𝐬𝐡 𝐭𝐡𝐚𝐭 $$(𝑓\circ 𝑔)^{(𝑛)}(𝑥)$$ 𝐞𝐱𝐢𝐬𝐭𝐬 𝐟𝐨𝐫 𝐚𝐥𝐥 𝐱 𝐢𝐧 𝐬𝐨𝐦𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 𝐚𝐫𝐨𝐮𝐧𝐝 $$𝑎$$?

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 $$(f\circ g)^{(n)}$$.

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 $$a$$?

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