Taylor and Big-O

Question

Given:

$P_n(x)$ is the Taylor polynomial for $f(x)$ at $x=a$.

$Q_n(x)$ is a polynomial of degree at most $n$. Where $f(x)=Q_n(x)+O(x-a)^{n+1}$ as $x \to a$

Now my calculus-book states:

$R_n(x)=Q_n(x)-P_n(x)=O((x-a)^{n+1})$ as $x\to a$

By replacing $x$ with $a+(x-a)$ and expanding powers we can write $R_n(x)$ in the form $R_n(x)=c_0+c_1(x-a)+c_2(x-a)^2+...+c_n(x-a)^n$.

QUESTION:

• Is it correct to say that the order of both $Q_n(x)$ and $P_n(x)$ equals $O((x-a)^n)$ as $x \to a$?
• Then why is the order of $R_n(x)=Q_n(x)-P_n(x)=O((x-a)^{n+1})$ as $x\to a$?
• Doesn't the last statement in my calculusbook imply that $R_n(x)=O((x-a)^n)$ as $x \to a$?

Context of the question

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EDIT

After some thought I think my confusion came from the fact that $O$ is different when $x \to a$ as opposed to $x \to \infty$.

Answer 1

Is it correct to say that the order of both $Q_n(x)$ and $P_n(x)$ equals $O((x-a)^n)$ as $x \to a$?

No, the order of both equals $O(1)$ as $x\to a$ since near $a$ that is the dominant degree.

Then why is the order of $R_n(x)=Q_n(x)-P_n(x)=O((x-a)^{n+1})$ as $x\to a$?

Because if you substract these two equations: $$f(x)=P_n(x)+O((x-a)^{n+1})$$ $$f(x)=Q_n(x)+O((x-a)^{n+1})$$ You'll get: $$0=P_n(x)-Q_n(x)+O((x-a)^{n+1})$$ $$Q_n(x)-P_n(x)=O((x-a)^{n+1})$$

Doesn't the last statement in my calculusbook imply that $R_n(x)=O((x-a)^n)$ as $x \to a$?

No, the order of $R_n(x)=c_0+c_1(x-a)+c_2(x-a)^2+...+c_n(x-a)^n$ equals $O(1)$ as $x\to a$

Answer 2

It is saying that $$ R_n(a+h)=c_0+c_1h+...+c_nh^n $$ the degree is bound by $n$ since both $P_n$ and $Q_n$ have degree $n$.

In the end the conclusion should be $R_n=0$ since all the $c_k=0$. If not, then for the smallest $k$ with $c_k\ne 0$ you would get $R_n(x)=O((x-a)^k)$ for some $k\le n$ contradicting the assumptions.