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


Give a more direct proof by writing



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


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




we have

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


Part $(c)$ shows that


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 $>n$.


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

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.

1 Answer 1


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.

  • $\begingroup$ 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 $\frac{f^{(n+1)}(t)}{(n+1)!}(x-b)^{n+1}$? $\endgroup$
    – xoux
    Commented Oct 12, 2022 at 7:35
  • 1
    $\begingroup$ 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. $\endgroup$
    – grand_chat
    Commented Oct 12, 2022 at 16:07
  • 3
    $\begingroup$ 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 $x$. $\endgroup$
    – grand_chat
    Commented Oct 12, 2022 at 16:13
  • 1
    $\begingroup$ 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. $\endgroup$
    – grand_chat
    Commented Oct 13, 2022 at 16:14
  • $\begingroup$ 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 $\displaystyle\lim_{x\to a}\frac{f \circ g (x)}{(x-a)^n}=\lim_{x \to a}\frac{P_{n,b,f} \circ P_{n,a,g}(x)}{(x-a)^n}$ $\endgroup$
    – S.C.
    Commented Jan 30 at 3:00

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