Understanding Theorem 13 from Calculus 7th ed, R. Adams, C. Essex, 4.10

[mcastillo356]

TL;DR Summary

I'm sweating to understand it. There is a reasoning, under an assumption, that it will happen to be contradictory that some polynomial is not identically zero. The algebra is difficult to me.

The following properties of big-O notation follow from the definition:

(i) if $f(x)=O(u(x))$ as $x\rightarrow{a}$, then $Cf(x)=O(u(x))$ as $x\rightarrow{a}$ for any value of the constant $C$.
(ii) If $f(x)=O(u(x))$ as $x\rightarrow{a}$ and $g(x)=O(u(x))$ as $x\rightarrow{a}$, then $f(x)\pm{g(x)}=O(u(x))$ as $x\rightarrow{a}$.
(iii) If $f(x)=O((x-a)^{k}u(x))$ as $x\rightarrow{a}$, then $f(x)/(x-a)^{k}=O(u(x))$ as $x\rightarrow{a}$ for any constant $k$.

Taylor's Theorem says that if $f^{(n+1)}(t)$ exists on an interval containing $a$ and $x$, and $P_n$ is the $n$th-order Taylor polynomial for $f$ at $a$, then, as $x\rightarrow{a}$

$$f(x)=P_{n}(x)+O\Big((x-a)^{n+1}\Big)$$

This is a statement about how rapidly the graph of the Taylor polynomial $P_{n}(x)$ approaches that of $f(x)$ as $x\rightarrow{a}$; the vertical distance between the graph decreases as fast as $|x-a|^{n+1}$. The following theorem shows that the Taylor polynomial $P_{n}(x)$ is the only polynomial of degree at most $n$ whose graph approximates the graph of $f(x)$ that rapidly

Theorem 13

If $f(x)=Q_{n}(x)+O((x−a)n+1)$ as $x→a$ where $Q_n$ is a polynomial of degree at most $n$, then $Q_{n}(x)=P_{n}(x)$, that is, $Q_n$ is the Taylor polynomial for $f(x)$ at $x=a$.

PROOF

Let $P_n$ be the Taylor polynomial, then properties (i) and (ii) of Big-O imply that $$R_{n}(x)=Q_{n}(x)−P_{n}(x)=O((x−a)^{(n+1})$$
We want to show that $R_{n}(x)$ is identically zero so that $Q_n(x)=P_n(x)$ for all $x$. By replacing $x$ with $a+(x-a)$ and expanding powers, we can write $R_n(x)$ in the form $$R_{n}(x)=c_0+c_1(x-a)+c_2(x-a)^2+\ldots+c_n(x-a)^n$$. If $R_n(x)$ is not identically zero, then there is a smalest coefficient $c_k (k\leq n)$, such that $c_k\neq 0$, but $c_j=0$ for $0\leq j\leq {k-1}$, so $c_k(x-a)^k+c_{k+1}(x-a)^{k+1}+\ldots +c_n(x-a)^{n-k}$. Therefore $\lim_{x\rightarrow{a}}R_{n}(x)/(x-a)=c_k\neq 0$. However, by property (iii) above we have $R_{n}(x)/(x-a)^k=O((x-a)^{n+1-k})$. Since $n+1-k>0$, this says $R_{n}(x)/(x-a)^{k}\rightarrow{0}$ as $x\rightarrow{a}$. This contradiction shows that $R_{n}(x)$ must be identically zero. Therefore $Q_{n}(x)=P_{n}(x)$ for all $x$.

The real question is if should I face this theorem naively, or seriosly; if I take it easy, I understand it. If I get obstinate, and want to check right the theorem, then there is a section for which I only got a mess:

How emerges $R_{n}(x)=(x-a)^k(c_k+c_{k+1}(x-a)+\ldots{+c_n(x-a)^{n-k}})\Rightarrow{R_{n}(x)=(x-a)^k c_k+c_{k+1}(x-a)^{k+1}+\ldots+ c_n(x-a)^n}$?

Because if there is a minimum $c_k\neq 0(k\leq n)$; this coefficient will be zero if $0\leq j\leq {k-1}$, this is, in the closed interval $[0,k-1]$, $R_{n}(x)=(x-a)^k c_k+c_{k+1}(x-a)^{k+1}+\ldots+ c_n(x-a)^n$. Just factorize

 

[hutchphd]

TL;DR Summary: I'm sweating to understand it. There is a reasoning, under an assumption, that it will happen to be contradictory that some polynomial is not identically zero. The algebra is difficult to me.

The following properties of big-O notation follow from the definition:

(i) if $f(x)=O(u(x))$ as $x\rightarrow{a}$, then $Cf(x)=O(u(x))$ as $x\rightarrow{a}$ for any value of the constant $C$.
(ii) If $f(x)=O(u(x))$ as $x\rightarrow{a}$ and $g(x)=O(u(x))$ as $x\rightarrow{a}$, then $f(x)\pm{g(x)}=O(u(x))$ as $x\rightarrow{a}$.
(iii) If $f(x)=O((x-a)^{k}u(x))$ as $x\rightarrow{a}$, then $f(x)/(x-a)^{k}=O(u(x))$ as $x\rightarrow{a}$ for any constant $k$.

Taylor's Theorem says that if $f^{(n+1)}(t)$ exists on an interval containing $a$ and $x$, and $P_n$ is the $n$th-order Taylor polynomial for $f$ at $a$, then, as $x\rightarrow{a}$

$$f(x)=P_{n}(x)+O\Big((x-a)^{n+1}\Big)$$

This is a statement about how rapidly the graph of the Taylor polynomial $P_{n}(x)$ approaches that of $f(x)$ as $x\rightarrow{a}$; the vertical distance between the graph decreases as fast as $|x-a|^{n+1}$. The following theorem shows that the Taylor polynomial $P_{n}(x)$ is the only polynomial of degree at most $n$ whose graph approximates the graph of $f(x)$ that rapidly

Theorem 13

If $f(x)=Q_{n}(x)+O((x−a)n+1)$ as $x→a$ where $Q_n$ is a polynomial of degree at most $n$, then $Q_{n}(x)=P_{n}(x)$, that is, $Q_n$ is the Taylor polynomial for $f(x)$ at $x=a$.

PROOF

Let $P_n$ be the Taylor polynomial, then properties (i) and (ii) of Big-O imply that $$R_{n}(x)=Q_{n}(x)−P_{n}(x)=O((x−a)^{(n+1})$$
We want to show that $R_{n}(x)$ is identically zero so that $Q_n(x)=P_n(x)$ for all $x$. By replacing $x$ with $a+(x-a)$ and expanding powers, we can write $R_n(x)$ in the form $$R_{n}(x)=c_0+c_1(x-a)+c_2(x-a)^2+\ldots+c_n(x-a)^n$$. If $R_n(x)$ is not identically zero, then there is a smalest coefficient $c_k (k\leq n)$, such that $c_k\neq 0$, but $c_j=0$ for $0\leq j\leq {k-1}$, so $c_k(x-a)^k+c_{k+1}(x-a)^{k+1}+\ldots +c_n(x-a)^{n-k}$. Therefore $\lim_{x\rightarrow{a}}R_{n}(x)/(x-a)=c_k\neq 0$. However, by property (iii) above we have $R_{n}(x)/(x-a)^k=O((x-a)^{n+1-k})$. Since $n+1-k>0$, this says $R_{n}(x)/(x-a)^{k}\rightarrow{0}$ as $x\rightarrow{a}$. This contradiction shows that $R_{n}(x)$ must be identically zero. Therefore $Q_{n}(x)=P_{n}(x)$ for all $x$.

The real question is if should I face this theorem naively, or seriosly; if I take it easy, I understand it. If I get obstinate, and want to check right the theorem, then there is a section for which I only got a mess:

How emerges $R_{n}(x)=(x-a)^k(c_k+c_{k+1}(x-a)+\ldots{+c_n(x-a)^{n-k}})\Rightarrow{R_{n}(x)=(x-a)^k c_k+c_{k+1}(x-a)^{k+1}+\ldots+ c_n(x-a)^n}$?

Because if there is a minimum $c_k\neq 0(k\leq n)$; this coefficient will be zero if $0\leq j\leq {k-1}$, this is, in the closed interval $[0,k-1]$, $R_{n}(x)=(x-a)^k c_k+c_{k+1}(x-a)^{k+1}+\ldots+ c_n(x-a)^n$. Just factorize

I believe $\lim_{x\rightarrow{a}}R_{n}(x)/(x-a)=c_k\neq 0$ should be $$\lim_{x\rightarrow{a}}R_{n}(x)/(x-a)^k=c_k\neq 0$$ and then the rest seems absolutely straightforward to me (and I am easily confused). Does that help?

 

[mcastillo356]

I believe $\lim_{x\rightarrow{a}}R_{n}(x)/(x-a)=c_k\neq 0$ should be $$\lim_{x\rightarrow{a}}R_{n}(x)/(x-a)^k=c_k\neq 0$$

Yes.

and then the rest seems absolutely straightforward to me (and I am easily confused). Does that help?

Let $n=3$
$R_{3}(x)=c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3$, if isn't identically zero, some $c_i\neq0$. If the smallest non-zero coefficient happens to be zero, $c_0\neq 0$, there cannot be factorization

 

[hutchphd]

$$(x-a)^0=1$$. Where does this fail?

 

[mcastillo356]

HI, @hutchphd, PF

$$(x-a)^0=1$$. Where does this fail?

It doesn't fail. Your post is been the kick I needed. P&L.

Marcos Castillo

PD: $c_j\neq 0$ for $0\leq j\leq {k-1}$. How do I harmonize what I perceive as a cognitive dissonance?

 

[mcastillo356]

Solved. If $c_j\neq 0$ for $0\leq j\leq{k-1}$, the textbook is very concrete: "If $R_{n}(x)$ is not identically zero, then there is a smalest coefficient $c_k(k\leq n)$ such that $c_k$...". If $k\leq n$, $0\leq\ldots k-3\leq k-2\leq k-1\leq k\leq n$. Therefore $c_j$ changes to $c_k$
Solved?

 

[mcastillo356]

"The real importance of Theorem 13 is that it enables us to obtain Taylor polynomials for new functions by combining others already known: as long as the error term - Of those known?- is of higher degree than the order of the polynomial obtained, the polynomial must be the Taylor polynomial."

Example Find the Maclaurin polynomial of order $2n$ for $\cosh x=\dfrac{e^{x}+e^{-x}}{2}$

The error term of $e^{-x}$ is of lower degree ($O(x^{2n-2})$) than the one of the Maclaurin polynomial $P_{2n}(x)$ for $\cosh x$: $O(x^{2n+2})$

There seems to be a contradiction

 

[hutchphd]

Why must they be equal? Q_n is a polynomial of degree at most n.

 

[mcastillo356]

$$f(x)=Q_{n}(x)+O((x-a)^{n+1})\underrightarrow{_{n\rightarrow{2n+1}}}f(x)=Q_{2n+1}(x)+O((x-a)^{2n+2})



\Rightarrow{e^x=Q_{2n+1}(x)+O((x-a)^{2n+2})}



\Rightarrow{e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+ \ldots+\dfrac{x^{2n}}{(2n)!}+\dfrac{x^{2n+1}}{(2n+1)!}+ O(x^{2n+2})}$$

$$\Leftrightarrow$$

$$f(x)=P_{n}(x)+O((x-a)^{n+1})\underrightarrow{_{n\rightarrow{2n+1}}}f(x)=P_{2n+1}(x)+O((x-a)^{2n+2})



\Rightarrow{e^x=P_{2n+1}(x)+O((x-a)^{2n+2})}



\Rightarrow{e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+ \ldots+\dfrac{x^{2n}}{(2n)!}+\dfrac{x^{2n+1}}{(2n+1)!}+ O(x^{2n+2})}$$

Am I well on the track? Does it observe Theorem 13 and why?

 

[mcastillo356]

Hi, PF, I think I'm on the track

Why must they be equal? Q_n is a polynomial of degree at most n.

This is the key: $Q_n$ is at most a polynomial of degree $n$. The case where obtains the $2n$-th degree polynomial for $\cosh x$ is a based on a simple substitution, $n\rightarrow{2n+1}$. No violation of Theorem 13 of the textbook. Carefull and good sense's observance instead.

The real importance of Theorem 13 is that inables us to obtain Taylor polynomials for new functions by combining others already known; as the error term is of higher degree than the order of the polynomial obtained, the polynomial must be the Taylor polynomial.

The sentence in bold typing of the above quote is, in my dilettante opinion, of poor interest. I've tried to find the ephitet to describe it, but no word fits perfectly; redundant, futile, irrelevant, recurrent, resonant; once experienced Taylor's Theorem.

Hope this post written on my own is valid

 

[mcastillo356]

The sentence in bold typing of the above quote is, in my dilettante opinion, of poor interest. I've tried to find the ephitet to describe it, but no word fits perfectly; redundant, futile, irrelevant, recurrent, resonant; once experienced Taylor's Theorem.

Hi, PF,Theorem13 has nothing to see with Taylor's Theorem. It's just algebra.

P&L, Marcos