Applying the induction hypothesis indirectly? I have trouble understanding this proof. I am confused about the mechanics of the following proof (page 423, chapter 20, of the fourth edition of Spivak's Calculus (Taylor's Theorem)):

I am not sure that I understand how applying the induction hypothesis to $R'$ works in proving the lemma for $R$ for all $n$. I expected that the induction hypothesis would be applied to $R$. 
Maybe my confusion stems from the fact that I don't apply the induction hypothesis generally enough. Maybe I need to assume that it holds for some $n$ for all functions $f$ that meet the given criteria and that it therefore also holds for $f'$, from which the case $n+1$ follows for all $f$. So we assume that the induction hypothesis is true for both $R$ and $R'$ but use the latter to derive the $n+1$ case for the former. This also requires that $R'$ is $(n)$-times differentiable with $R'^{(k)}(a) = 0$ for all k up to $n-1$. But this appears always to be the case. 
 A: Yes, the statement $P(n)$ that you prove is

For every $n+1$ times differentiable function $R$, ...

and has only $n$ as free variable. Everything else, in particular $R$ is quantified within $P(n)$. 
For induction, you show $P(0)$, i.e.,

For every (once) differentiable function $R$, ...

and you show $\forall n\colon P(n)\to P(n+1)$ to finally arrive at

For all $n\in\Bbb N_0$, for every $n+1$ times differentiable function $R$ ...

For the induction step $\forall n\colon P(n)\to P(n+1)$, we proceed as follows:
Let $n\in\Bbb N_0$ be arbitrary. We want to show $P(n)\to P(n+1)$. 
So assume $P(n)$. Now we want to show $P(n+1)$. So let $R$ be an arbitrary $n+2$ times differentiable with $R(a)=\ldots=R^{(n+1)}(a)=0$. Then $R'$ is $n+1$ times differentiable with $(R')'(a)=\ldots=(R')^{(n)}(a)=0$, hence $P(n)$ applies to $R'$ ... ... hence the conclusion holds for $R$. As $R$ was arbitrary, we conclude that $P(n+1)$ holds. Thus we have shows $P(n)\to P(n+1)$. As $n$ was arbitrary, we conclude $\forall n\in\Bbb N_0\colon P(n)\to P(n+1)$.
A: We have taken the function  $R $ such that it is  $n+2$ times differentiable and $R^{(k)}(a)=0$,$k=0,1,\ldots,n+1$. Our Induction assumption is that the proposition holds for for all functions with degree less than that of $R$. Now if we take $R'$, it has one degree less than that of $R$, it is $(n+1) $ times differentiable and since $R^{(k)}(a)=0$,
$k=0,1,\ldots,n+1$, we have $(R')^{(k)}(a)=0$,
$k=0,1,\ldots,n$. Hence we can apply the induction assumption for $R'$ and proceed as shown in the textbook. 
