Spivak, Ch. 20: Understanding a step in the proof of lemma

[zenterix]

TL;DR Summary

In Chapter 20, "Approximation by Polynomial Functions", there is a lemma (used to prove Taylor's Theorem right after), the proof of which includes a step that I would like to understand better.

In Chapter 20 of Spivak's Calculus is the lemma shown below (used afterward to prove Taylor's Theorem). My question is about a step in the proof of this lemma.

Lemma: Suppose that the function $R$ is $(n+1)$-times differentiable on $[a,b]$ and

$$R^{(k)}(a)=0, \text{ for } k=0,1,2,...,n$$

Then for any $x$ in $(a,b]$ we have

$$\frac{R(x)}{(x-a)^{n+1}}=\frac{R^{(n+1)}(t)}{(n+1)!}, \text{ for
some } t \text{ in } (a,x)\tag{1}$$

Here is the proof as it appears in the book

For $n=0$ this is just the Mean Value Theorem, and we will prove the
theorem for all $n$ by induction on $n$. To do this we use the Cauchy
Mean Value Theorem to write

$$\frac{R(x)}{(x-a)^{n+2}}=\frac{R'(z)}{(n+2)(z-a)^{n+1}}=\frac{1}{n+2}\frac{R'(z)}{(z-a)^{n+1}},
\text{ for some } z \text{ in } (a,x)$$

and then apply the induction hypothesis to $R'$ on the interval
$[a,z]$ to get

$$\frac{R(x)}{(x-a)^{n+2}}=\frac{1}{n+2}\frac{(R')^{(n+1)}(t)}{(n+1)!},
\text{ for some } t \text{ in } (a,z)\tag{2}$$

$$=\frac{R^{(n+2)}(t)}{(n+2)!}$$

My question is: how do we know that $(R')^{n+1}$ is defined in $(2)$?

Let me try to go through the proof in more steps.

For $n=0$ we have

$$\frac{R(x)}{x-a}=\frac{R(x)-R(a)}{x-a}=R'(t)$$

for some $t\in (a,x)$, which is the MVT applied to $R$ on $[a,x]$.

Now assume as our induction hypothesis that $(1)$ is true for some $k$.

First, let's apply Cauchy MVT to the functions $R(x)$ and $g(y)=(y-a)^{k+2}$.

$$\frac{R(x)-R(a)}{g(x)-g(a)}=\frac{R'(z)}{g'(z)}$$

for some $z\in (a,x)$. Then

$$\frac{R(x)}{g(x)}=\frac{R(x)}{(x-a)^{k+2}}=\frac{R'(z)}{(k+2)(z-a)^{k+1}}=\frac{1}{k+2}\frac{R'(z)}{(z-a)^{k+1}}$$

At this point, we are to apply the induction hypothesis to $R'$ on $[a,z]$. In the induction hypothesis, $R$ is assumed to be $(n+1)$-times differentiable. Doesn't this mean that $R'$ is only $n$-times differentiable?

How do we know that $(R')^{n+1}$ is defined in $(2)$?

 

[zenterix]

Looks like the answer is a silly oversight.

Since the induction is over $n$, this includes the part about the function $R$ being $(n+1)$-times differentiable.

After the inductive hypothesis, we want to prove the result for some $k+1$. To do this we assume that the function $R$ is $(n+2)$-times differentiable.