# Spivak, Ch. 20, Problem 9d: Understanding Solution Manual Proof

The following problem is from Chapter 20 of Spivak's Calculus, "Approximation by Polynomial Functions".

My question is about item $$(d)$$, and I have previously asked a question about the comment at the end of item $$(c)$$.

1. (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 $$>n$$.

(d) If $$a=0$$ and $$b=g(a)=0$$, then

$$P_{n,a,f\circ g}=[P_{n,b,f}\circ P_{n,a,g}]_n$$

Here is what the solution manual says

Writing

$$f(x)=P_{n,0,f}(x)+R_{n,0,f}(x)$$

$$g(x)=P_{n,0,g}(x)+R_{n,0,g}(x)$$

we have

$$(f\circ g)(x)=P_{n,0,f}(P_{n,0,g}(x)+R_{n,0,g}(x))+R_{n,0,f}(g(x))$$

$$=A+B$$

Part $$(c)$$ shows that

$$A=P_{n,0,f}(P_{n,0,g}(x))+\bar{R}(x)$$

where $$\lim\limits_{x\to a} \frac{\bar{R}(x)}{(x-a)^n}=0\tag{3}$$ and the remark added at the end of $$(c)$$ shows that $$\lim\limits_{x\to a} \frac{B}{(x-a)^n}=0\tag{4}$$

Then, applying $$(c)$$ once again, we have

$$(f\circ g)(x)=(P_{n,0,f}\circ P_{n,0,g})(x)+\bar{\bar{R}}(x)\tag{5}$$

where $$\lim\limits_{x\to a} \frac{\bar{\bar{R}}(x)}{(x-a)^n}=0$$. It follows, just as in part $$(b)$$, that $$P_{n,0,f\circ g}=[P_{n,0,f}\circ P_{n,0,g}]_n$$

Everything up to $$(3)$$ is fine. How do we obtain $$(4)$$?

My attempt at understanding it is:

Since $$R_{n,0,f}$$ is polynomial in $$x-a=x$$, composed of a single term of degree $$n+1$$, and $$g(x-a)=g(x)$$ is such that the constant term in its Taylor polynomial is zero ($$g(a)=0$$ by assumption), then as per the comment at the end of $$(c)$$ all of the terms in $$R_{n,0,f}(g(x))$$ are of degree $$>n$$.

Hence

$$\lim\limits_{x\to a} \frac{R_{n,0,f}(g(x))}{(x-a)^n}=0$$

My other question is: how exactly is part $$(c)$$ applied again to reach $$(5)$$? Ie, in the context of part $$(c)$$, when are the polynomials $$p$$ and $$q$$ here in part $$(d)$$?

EDIT: Clarifications required by the bounty

• The solution manual sets $$a=b=g(a)=0$$. I'd like to see a proof of $$(d)$$ that does not require $$a=0$$ or $$b=0$$, only $$b=g(a)$$.
• I will write an answer specifying my current understanding of the problem and solution.

As a comment on your attempt, note that $$R_{n,b,f}$$ is not a polynomial. However, by hypothesis it can be written $$R_{n,b,f}(x)=(x-b)^n r_{n,b,f}(x)$$ where $$\lim \limits_{x\to b} r_{n,b,f}(x)=0$$.

The proof of (4) involves the corrected versions of both (c) and the remark at the end of (c). It doesn't require $$a=0$$ or $$b=0$$ but does assume $$b=g(a)$$.

Substitute $$g(x)$$ into $$R_{n,b,f}(x) = (x-b)^n r_{n,b,f}(x)$$ to obtain $$B:=R_{n,b,f}(g(x)) = [g(x)-b]^n r_{n,b,f}(g(x)).$$ We can write $$[g(x)-b]^n=\big[P_{n,a,g}(x) + R_{n,a,g}(x)-b\big]^n = p\big(q(x) + R_{n,a,g}(x)\big)$$ where $$p$$ and $$q$$ are the polynomials $$p(x):=[x-b]^n,\qquad q(x):=P_{n,a,g}(x).$$ Since $$R_{n,a,g}(x)/(x-a)^n\to0$$ as $$x\to a$$, we can apply (c) to get $$[g(x)-b]^n=p(q(x)) + \overline R(x)$$ where $$\lim\limits_{x\to a}\frac{\overline R(x)}{(x-a)^n}=0$$. Note now that $$q$$ is a polynomial in $$x-a$$, while $$p$$ is a polynomial in $$x-b$$ with $$b:=g(a)=P_{n,a,g}(a)=q(a)$$. So by the remark, $$p(q)$$ is a polynomial in $$x-a$$ with degree at least $$n$$.

Divide $$B$$ by $$(x-a)^n$$: $$\frac B{(x-a)^n} = \left[ \frac{p(q(x))}{(x-a)^n} + \frac{\overline R(x)}{(x-a)^n} \right]r_{n,b,f}(g(x))$$ As $$x\to a$$, the first term in square brackets tends to a constant, the second term tends to zero, and $$r_{n,b,f}(g(x))$$ tends to zero because $$g(x)$$ tends to $$g(a)=b$$. This proves (4).

As for (5), there's no need to invoke (c) again. The argument establishes two Taylor polynomials of degree $$n$$ for $$f\circ g$$ at $$a$$, both of whose remainders, when divided by $$(x-a)^n$$, tend to zero as $$x\to a$$, hence the two polynomials are equal.

• Why is $R_{n,b,f}$ not a polynomial? It is the remainder term of an application of Taylor's theorem to $f$. Is it not $\frac{f^{(n+1)}(t)}{(n+1)!}(x-b)^{n+1}$?
– xoux
Commented Oct 12, 2022 at 7:35
• Because $f$ in general is not a polynomial. The remainder $R_{n,b,f}$ is the difference between $f$ and the Taylor approximation of order $n$, which is a polynomial. If the remainder is a polynomial, then $f$ must be a polynomial. Commented Oct 12, 2022 at 16:07
• To clarify the meaning of the remainder term, it is an expression valid for each $x$. While $(x-b)^{n+1}$ is a polynomial in $x$, the $t$ that appears in the "coefficient" $\frac{f^{(n+1)}(t)}{(n+1)!}$ depends on $x$. Thus the coefficient is actually a function of $x$. Commented Oct 12, 2022 at 16:13
• I just noticed that you are seeking clarifications. The answer that I posted involves $a$ and $b$ but does not assume $a=0$ or $b=0$. It uses only the assumption $g(a)=b$, in two places. Commented Oct 13, 2022 at 16:14
• Great proof. For anyone going through this, note that $r_{n,b,f}(x)$ is implicitly defined as $\frac{f^{(n+1)}(t)}{(n+1)!}(x−b)$. Additionally, matching up the solution manual's corrected claim with the terms used here, the penultimate step would like $\displaystyle\lim_{x\to a}\frac{f \circ g (x)}{(x-a)^n}=\lim_{x \to a}\frac{P_{n,b,f} \circ P_{n,a,g}(x)}{(x-a)^n}$
– S.C.
Commented Jan 30 at 3:00