Exercise 3 in Section 7.11 of Apostol's Calculus (Vol. 1) little o-notation

Question

The problem is stated as:

Find the polynomial $P(x)$ of minimal degree such that $$ \sin(x-x^2)=P(x)+o(x^6)\quad \textrm{as }x\to 0. $$

and the answer in the book is

$$ P(x)=x-x^2-\frac{x^3}{6}+\frac{x^4}{2}-\frac{59x^5}{120}+\frac{x^6}{8} $$

I am wondering why the answer cannot be

$P(x) = x-x^2$

Since we can use the fact that $\sin(x-x^2) = x-x^2 + o((x-x^2)^2)$ as $x \to 0$

And we know that $o((x-x^2)^2) = o(x^5) = o(x^6)$ as $x$ goes to zero.

Then why is the answer from the back of the book not equal to $x-x^2$??

Answer 1

Note that one has $o(x^6)=o(x^5)$ (as $x\to 0$), but not $o(x^5)=o(x^6)$.

Also, $o((x-x^2)^2)=o(x^5)$ is incorrect: one has $\lim_{x\to 0}\frac{x^5}{(x-x^2)^2}=0$ but $\lim_{x\to 0}\frac{x^5}{x^5}=1$. In general, $o(f(x))=o(g(x))$ does NOT imply that $o(g(x))=o(f(x))$ as discussed in an answer to another question of yours.

Why the answer cannot be $x-x^2$?

It is an instructive exercise to show (using the correct answer in Apostol) that the following is FALSE: $$ \lim_{x\to 0}\frac{\sin (x-x^2)-(x-x^2)}{x^6}= 0\tag{1} $$

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[Added.] To see why (1) is false, you can use the solution of $P(x)$ in the book so that $$\begin{align} f(x):=\frac{\sin (x-x^2)-(x-x^2)}{x^6} &=\frac{a_3x^3+a_4x^4+a_5x^5+a_6x^6+o(x^6)}{x^6}\\ &=\frac{1}{x^3} \left( a_3+a_4x+a_5x^2+a_6x^3 \right)+\frac{o(x^6)}{x^6}\\ &=\frac{1}{x^3}\cdot g(x)+h(x) \end{align} $$ where $g(x):=a_3+a_4x+a_5x^2+a_6x^3$.

Now, note that $$ \lim_{x\to 0}g(x)=a_3\neq 0,\quad \lim_{x\to 0}\frac{o(x^6)}{x^6}=0. $$ So $ \lim_{x\to 0+}f(x)=\infty $ and (1) is thus impossible.

Answer 2

What they are asking you is to find a polynomial $p(x)$ such that $$ \lim_{x\rightarrow0}\frac{|\sin(x-x^2)-p(x)|}{x^6}=0 $$

No to your statements:

1. $o((x-x^2)^2) =o(x^5)$ is in general false. For example $h(x)=x(x-x^2)^2=o((x-x^2)^2$ as $x\rightarrow0$ but $x^{-5}h(x)\nrightarrow0$ as $x\rightarrow0$.
2. $o(x^5) = o(x^6)$ is false in general: $g(x):=x^6=o(x^5)$ but $g(x) \neq o(x^6)$.

To help you solve the problem, use Taylor expansion of $\sin z= z-\frac{z^3}{6}+\frac{z^5}{5!}-\frac{z^7}{7!}+...$. Notice that when $z=x-x^2$, $z^7$ is a polynomial on $x$ of degree 14 of the $z^7=x^7-7x^8+...-x^{14}$ This term is of course $o(x^6)$ since all powers involved are larger than $6$. So truncating the sine series up to the $7-th$ power gives the right decay of residual. The rest is to check that the $$ (x-x^2)-\frac{(x-x^2)^3}{6}+\frac{(x-x^2)^5}{5!} $$ is the polynomial in the statement of your problem. A little but tedious algebra.

Answer 3

A sketch of the correct method:

Since $x-x^2$ has order $1$, you have to expand $\sin u$ up to degree $6$, and substitute $x-x^2$ in $u, u^2,\dots u^6$, truncating the expansions of the different $(x-x^2)^k$ at degree $6$, which leads to a recursive computation: \begin{align} (x-x^2)^2&=x^2-2x^3+x^4& \quad(x-x^2)^3&=x^3-3x^4+3x^5-x^6 \\ (x-x^2)^4&=x^4-4x^5+6x^6-4x^7+x^8&(x-x^2)^5&=(x-x^2)^4(x-x^2)\\ &=x^4-4x^5+6x^6+o(x^6)&&=\bigl((x^4-4x^5+6x^6+o(x^6)\bigr)(x-x^2)\\ (x-x^2)^6&=(x-x^2)^5(x-x^2)&&=\begin{aligned}[t]&x^5-4x^6+6x^7+o(x^7)\\&\phantom{x5}-x^6+4x^7-6x^8+o(x^8) &\end{aligned}\\&=\bigl(x^5-5x^6+o(x^6)\bigr)(x-x^2)&&=x^5-5x^6+o(x^6)\\ &=\dots=x^6+o(x^6) \end{align}

Can you end the computations?