Spivak, Ch. 20, Problem 9c: Understanding comment at end of problem item. 


*(a) Problem $7(i)$ amounts to the equation

$$P_{n,a,f+g}=P_{n,a,f}+P_{n,a,g}$$
Give a more direct proof by writing
$$f(x)=P_{n,a,f}(x)+R_{n,a,f}(x)\tag{1}$$
$$g(x)=P_{n,a,g}(x)+R_{n,a,g}(x)\tag{2}$$
and using the obvious fact about $R_{n,a,f}+R_{n,a,g}$.
(b) Similarly, Problem $7(ii)$ could be used to show that
$$P_{n,a,fg}=[P_{n,a,f}\cdot P_{n,a,g}]_n$$
where $[P]_n$ denotes the truncation of $P$ to degree $n$, the sum of
all terms of $P$ of degree $\leq n$ [with $P$ written as a polynomial
in $x-a$]. Again, give a more direct proof, using the obvious facts
about products involving terms of the form $R_n$.
(c) Prove that if $p$ and $q$ are polynomials in $x-a$ and
$\lim\limits_{x\to 0} \frac{R(x)}{(x-a)^n}=0$ then
$$p(q(x)+R(x))=p(q(x))+\bar{R}(x)$$
where $$\lim\limits_{x\to 0} \frac{\bar{R}(x)}{(x-a)^n}=0$$
Also note that if $p$ is a polynomial in $x-a$ having only terms of
degree $>n$, and $q$ is a polynomial in $x-a$ whose constant term is
$> 0$, then all terms of $p(q(x-a))$ are of degree $>n$.

My question regards the comment at the end of item $(c)$.
Let $p$ and $q$ be polynomials in $x-a$ such that terms in $p$ have degree $>n$, and the constant term in $q$ is $0$.
$$p(x)=\sum\limits_{i=n}^{m_1} a_i (x-a)^i$$
$$q(x)=\sum\limits_{i=1}^{m_2} b_i (x-a)^i$$
What is $q(x-a)$?
Is it
$$q(x-a)=\sum\limits_{i=1}^{m_2} b_i (x-2a)^i$$
Or is it just
$$q(x-a)=\sum\limits_{i=1}^{m_2} b_i (x-a)^i$$
In the first case we have
$$p(q(x-a))=\sum\limits_{i=n}^{m_1} a_i \left ( \sum\limits_{j=1}^{m_2} b_i (x-2a)^j -a \right )^i$$
In the second we have
$$p(q(x-a))=\sum\limits_{i=n}^{m_1} a_i \left ( \sum\limits_{j=1}^{m_2} b_i (x-a)^j -a \right )^i$$
Is this sort of what the comment is saying? If so, is the next step to somehow conclude that all $(x-a)$ factors have a degree $>n$ in this second expression?
 A: There are many typographical errors in part (c). Looking at the intended application of part (c) and its remark, the most general true statement is:

If $p$ and $q$ are polynomials and $\lim\limits_{x\to a}\frac{R(x)}{(x-a)^n}=0$ then
$$p(q(x) +R(x)) = p(q(x)) + \overline R(x)$$
where $$\lim_{x\to a}\frac{\overline R(x)}{(x-a)^n}=0.$$
Also note that if $q$ is a polynomial in $x-a$ and $p$ is a polynomial in $x-b$ with $b=q(a)$, then $p(q)$ is a polynomial in $x-a$. Moreover, if $p$ has only terms of degree $\ge n$, then all terms of $p(q)$ are of degree $\ge n$.

To prove (c), first write the Taylor polynomial for $p$ at $a$:
$$p(t) = \sum_{k=0}^n \frac{p^{(k)}(a)}{k!}(t-a)^k$$ Substitute $a=q(x)$ and $t=q(x)+R(x)$:
$$p(q(x)+R(x))=\sum_{k=0}^n \frac{p^{(k)}(q(x))}{k!}(R(x))^k
$$ Subtracting $p(q(x))$ starts the sum at $k=1$:
$$\overline R(x):=p(q(x)+R(x))-p(q(x))=\sum_{k=1}^n \frac{p^{(k)}(q(x))}{k!}(R(x))^k
$$
Divide through by $(x-a)^n$:
$$\frac{\overline R(x)}{(x-a)^n}:=\sum_{k=1}^n \frac{p^{(k)}(q(x))}{k!}\left[\frac{R(x)}{(x-a)^n}\right]^k(x-a)^{n(k-1)}
$$
Now let $x\to a$. The quantity in square brackets tends to zero, while every other factor in the sum tends to a constant (possibly zero).

The remark is easily proved: If $p(x)=\sum_{k=n}^N p_k(x-b)^k$ and $q(x)=\sum_{i=0}^Mq_i(x-a)^i$, then
$$
p(q(x)) = \sum_{k=n}^N p_k(q(x)-b)^k.
$$
But $b=q(a)=q_0$ so $$q(x)-b=q(x)-q_0=\sum_{i=1}^Mq_i(x-a)^i$$ is a polynomial in $x-a$ whose terms $(x-a)^i$ all have degree at least $1$. Therefore $p(q(x))$ is a polynomial in $x-a$, all of whose terms have degree at least $n$.
