- #1

mcastillo356

Gold Member

- 560

- 267

- 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) 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