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

In summary, we discussed the properties of big-O notation and how they apply to Taylor's Theorem. The theorem states that if a function ##f(x)## can be approximated by a polynomial ##Q_n(x)## of degree at most ##n##, then the polynomial must be the Taylor polynomial ##P_n(x)## for ##f(x)## at a specific point ##x=a##. This is proven by showing that the difference between the two polynomials, ##R_n(x)##, must be identically zero, or else it would contradict the properties of big-O notation. The only potential confusion arises when we try to expand ##R_n(x)##, but it can be easily resolved by factoring
  • #1
mcastillo356
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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
 
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  • #2
mcastillo356 said:
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?
 
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  • #3
hutchphd said:
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.

hutchphd said:
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
 
  • #4
$$(x-a)^0=1$$. Where does this fail?
 
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  • #5
HI, @hutchphd, PF

hutchphd said:
$$(x-a)^0=1$$. 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?
 
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  • #6
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?
 
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  • #7
"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
 
  • #8
Why must they be equal? Qn is a polynomial of degree at most n.
 
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  • #9
$$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?
 
  • #10
Hi, PF, I think I'm on the track
hutchphd said:
Why must they be equal? Qn 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
 
  • #11
mcastillo356 said:
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
 

What is Theorem 13 from Calculus 7th ed?

Theorem 13 is a fundamental result in calculus that states the Mean Value Theorem for Integrals. It connects the average value of a function over an interval to its definite integral.

What is the significance of Theorem 13?

Theorem 13 is significant because it allows us to find the average value of a function over an interval, which has many practical applications in fields such as physics, engineering, and economics.

How is Theorem 13 applied in calculus?

Theorem 13 is applied in calculus by using it to evaluate definite integrals. It allows us to calculate the average value of a function over an interval by finding the definite integral of the function over that interval.

What are the conditions for Theorem 13 to hold?

Theorem 13 holds when the function is continuous on the closed interval and differentiable on the open interval. It also requires that the endpoints of the interval have the same function value.

Can Theorem 13 be extended to higher dimensions?

Yes, Theorem 13 can be extended to higher dimensions through the Multivariable Mean Value Theorem. This theorem states that if a function is continuous on a closed and bounded region in n-dimensional space and differentiable on the interior of that region, then there exists a point where the gradient of the function is equal to the average value of the function over the region.

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