B Why is Big-O about how rapidly the Taylor graph approaches that of f(x)?

[mcastillo356]

TL;DR Summary

Don't understand Big-O Notation mathematical meaning

Hi, PF
For example, $\sin{x}=O(x)$ as $x\rightarrow{0}$ because $|\sin{x}|\leq{|x|}$ near 0. This fits textbook definition; easy, I think.
But, Taylor's Theorem says that if $f^{(n+1)}(t)$ exists on an interval containing $a$ and $x$, and if $P_{n}$ is the $n$th-order Taylor polynomial for $f$ at $a$, then, as $x\rightarrow{a}$,
$$f(x)=P_{n}(x)+O\Big({(x-a)^{n+1}}\Big)$$
and "this is a statement about how rapidly the graph of the Taylor polynomial $P_{n}(x)$ aproaches that of $f(x)$ as $x\rightarrow{a}$; the vertical distance between the graphs decreases as fast as $|x-a|^{n+1}$.
Italic words is what I don't understand
My attempt is somehow to restrict it to one example: MacLaurin polynomial for $e^x$:
$$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\ldots{+\dfrac{x^n}{n!}+O(x^{n+1})}$$
Why the graph decreases as fast as $|x|^{n+1}$?
Thanks! Hope LaTeX is right and message is understandable (post without preview, sorry).

 

[mcastillo356]

Sorry, I just got it: I must only substitute $x$ and $n$ ... Well, let's see, suppose ...If I look at the graph, and I want to know the value of $e^{1}$ for $P_{4}$, I get the sequence, I mean, the Maclaurin formula with errors in Big-O form...Let me think it over, sleep on it.

 

[mcastillo356]

Hi PF, working on it. I think must make an effort I haven't made still.
Greetings

 

[pasmith]

This follows from any of the explicit expressions for the remainder. For example, if f(x)<br /> = f(a) + (x - a)f&#039;(a) + \dots + \frac{(x - a)^n}{n!}f^{(n)}(a) + R_n(x) then by applying the mean value theorem to F(t) = f(t) + (x - t)f&#039;(t) + \dots + \frac{(x - t)^n}{n!} f^{(n)}(t) it follows that there exists \xi between a and x such that <br /> \begin{split}<br /> R_n(x) &amp;= F(x) - F(a) \\<br /> &amp;= (x- a)F&#039;(\xi) \\<br /> &amp;= <br /> (x - a) \frac{(x - \xi)^{n}}{n!}f^{(n+1)}(\xi).\end{split}

 

[mcastillo356]

Hi, PF, @pasmith, not familiar to $\xi$: been trying, though. There goes my attempt. First I quote the texbook, and then my point of view:

"Big-O Notation

We write $f(x)=O(u(x))$ as $x\rightarrow{a}$ provided that
$|f(x)|\leq{k|u(x)|}$
holds for some constant $k$ on some open interval containing $x=a$.
Similarly, $f(x)=g(x)+O(u(x))$ as $x\rightarrow{a}$ if $f(x)-g(x)=O(u(x))$ as $x\rightarrow{a}$, that is, if $|f(x)-g(x)|\leq{k|u(x)|}$ near $a$.
For example, $\sin{\;x}=O(x)$ as $x\rightarrow{0}$ because $|\sin{\;x}|\leq{|x|}$ near 0
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 if $P_{n}$ is the $n$th-order Taylor polynomial for $f$ at $a$, then, as $x\rightarrow{a}$,

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

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 graphs decreases as fast as $|x-a|^{n+1}$."

This precalculus quote is been useful to me. Eventually, in a simpler way, I think I've come to understand the quote, and outline the answer to my question on the first post of the thread: why the distance between graphs getting close to $f(x)$ decreases as fast as $|x-a|^{n+1}$?:

$f(x)=P_{n}(x)+O((x-a)^{n+1})$
$\Rightarrow{|f(x)-P_{n}(x)|\leq{k|(x-a)^{n+1}|}}$

where $k=\dfrac{E_{n}(x)}{(x-a)^{n+1}}=\dfrac{f^{(n+1)}(s)}{(n+1)!}$, for some $s$ between $a$ and $x$

Edited, reason: weird LaTeX used. Check it, PF, please.

 

[Ibix]

weird LaTeX used.

"You can't use 'macro parameter character #' in math mode" means you used a single # character to end some inline maths instead of two. It will be around where your post starts to look wierd.

 

[mcastillo356]

Hi, PF, @pasmith: I've been studying, trying to understand your post, and I don´t know if I am right: @pasmith , your post is about Taylor's Theorem with Cauchy's Remainder?
https://tartarus.org/gareth/maths/Analysis_1/Taylors_Theorem.pdf
If so, I would like to ask: why is the key to understand this thread's question? I think that I am bitting off more than I can chew, I mean I'm full of doubts.
My decission is to look for a math teacher, here where I live at. It is the wiser option.
But the main question is: Is it Taylor's Theorem with Cauchy's Remainder?
Thanks. Excuse my poor English

 

[mcastillo356]

Hi, PF

This follows from any of the explicit expressions for the remainder. For example, if f(x)<br /> = f(a) + (x - a)f&#039;(a) + \dots + \frac{(x - a)^n}{n!}f^{(n)}(a) + R_n(x) then by applying the mean value theorem to F(t) = f(t) + (x - t)f&#039;(t) + \dots + \frac{(x - t)^n}{n!} f^{(n)}(t) it follows that there exists \xi between a and x such that <br /> \begin{split}<br /> R_n(x) &amp;= F(x) - F(a) \\<br /> &amp;= (x- a)F&#039;(\xi) \\<br /> &amp;=<br /> (x - a) \frac{(x - \xi)^{n}}{n!}f^{(n+1)}(\xi).\end{split}

Nice quote indeed. I've tried to understand it; prove it, actually. There it goes:

1st- Apply the Mean Value Theorem to $$F(t) = f(t) + (x - t)f'(t) + \dots + \frac{(x - t)^n}{n!} f^{(n)}(t)$$ in the interval $[a,x]$, or $[x,a]$ (depending on $a<x$ or $x<a$); chosen $a<x$, this is, $$F(x)-F(a)=F'(\xi)(x-a)$$
Tasks: $F(x)-F(a)$ is in fact the difference between the function and its Taylor $n$th-order polynomial?; next, compute $F'(\xi)$

(i) $F(a)$ is, indeed, $P_{n}(x)$, this is, the $n$-th order Taylor polynomial for $F(x)$ at $x=a$;
(ii) Compute $F'(\xi)$

Proof by induction

1-for $n=0$: $F'(t)=f'(t)=\dfrac{(x-t)^n}{n!}f^{(n+1)}(t)$.
2-for $n=1$: $F'(t)=f'(t)+(x-t)f''(t)-f'(t)=\dfrac{(x-t)^n}{n!}f^{(n+1)}(t)$

Inductive step

Suppose proven for $n$, and prove it for $n+1$:

We got

$F(t)=\displaystyle\sum_{k=0}^n\dfrac{(x-t)^k}{k!}f^{(k)}(t)$

$F(t)=\displaystyle\sum_{k=1}^{n+1}{\dfrac{(x-t)^{k}}{k!}f^{(k)}}(t)=\underbrace{\displaystyle\sum_{k=1}^{n}{\dfrac{(x-t)^{k}}{k!}f^{(k)}}(t)}_{*}+\dfrac{(x-t)^{n+1}}{(n+1)!}f^{(n+1)}(t)$

It is possible to apply the induction hypotesis on $(*)$ when differentiating. Renders

$F'(t)=\underbrace{\dfrac{(x-t)^n}{n!}f^{(n+1)}(t)}_*-\dfrac{(x-t)^n}{n!}f^{(n+1)}(t)+\dfrac{(x-t)^{n+1}}{(n+1)!}f^{(n+2)}(t)=\dfrac{(x-t)^{n+1}}{(n+1)!}f^{(n+2)}(t)$

Proven, See you later. Still got some doubts. Take this post like some kind of rough draft.

Greetings. Please check LaTeX, post without preview

Peace, love.

Marcos

 

[mcastillo356]

Hi, PF, hope not to be cumbersome, but I've come to the conclusion that this thread lacks of completeness...Well, this word might not be appropiate; I mean coherence, consistency:

Must be provided a proof for the mentioned three properties of the Big O notation;
There is a Taylor's Theorem quote that leads to Theorem 13, mentioned in this thread, but not justified; I mean explained in detail.

I'll give a trie in the next posts.

Peace and Love!

 

[mcastillo356]

"Big-O Notation

We write $f(x)=O(u(x))$ as $x\rightarrow{a}$ provided that
$|f(x)|\leq{k|u(x)|}$
holds for some constant $k$ on some open interval containing $x=a$.
Similarly, $f(x)=g(x)+O(u(x))$ as $x\rightarrow{a}$ if $f(x)-g(x)=O(u(x))$ as $x\rightarrow{a}$, that is, if $|f(x)-g(x)|\leq{k|u(x)|}$ near $a$.
For example, $\sin{\;x}=O(x)$ as $x\rightarrow{0}$ because $|\sin{\;x}|\leq{|x|}$ near 0
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 $C$.

(i)$Cf(x)=O(u(x))\Rightarrow{|Cf(x)|\leq{k|u(x)|}}\Rightarrow{|f(x)|\leq{k|u(x)|}\leq{\dfrac{k}{C}|u(x)|}}\Rightarrow{\dfrac{k}{C}>0}$

 

[mcastillo356]

(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 $x\rightarrow{a}$.

(ii)
Got $|f(x)|\leq{k|u(x)|}$ and $|g(x)|\leq{k'|u(x)|}$. Thus, $|f(x)\pm{g(x)}|\leq{|f(x)|+|g(x)|}\leq{(k+k')}|u(x)|$

 

[mcastillo356]

(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 $k$

$f(x)=O((x-a)^{k}(u(x))\Leftrightarrow{|f(x)|\leq{k|(x-a)^{k}u(x)|}}\Leftrightarrow{f(x)/(x-a)^{k}}=O(u(x))$

 

[WWGD]

Still, @mcastillo356 , notice Sinx is somewhat of a special case, as $|sinx|\leq 1$

 

[mcastillo356]

Still, @mcastillo356 , notice Sinx is somewhat of a special case, as $|sinx|\leq 1$

@WWGD, very interesting your mention, but what difference does it make? I mean, trigonometric inverse functions are not my aim. In fact $\sin{\;x}$ is just an example. The reason for it to appear is that helps introducing Big $O$ inmediately, just as $\mbox{Sin y}$. My aim is, at this first step, explore Big $O$; but the domain of the functions will remain at the abscissas.

 

[mcastillo356]

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

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 graphs decreases as fast as $|x-a|^{n+1}$.

I don't understand this quote. My attempt:

$f(x)-P_{n}(x)=O\Big(|x-a|^{n+1}\Big)$. On the right side there is the error term, the rapidness Taylor's polynomial approaches to $f(x)$, and third:
$$|f(x)-P_{n}(x)|\leq{k|x-a|^{n+1}}$$
That means it exists a constant $k$ such that the distance between the function and the Taylor polynomial is bounded by a multiple of the function $|x-a|^{n+1}$. I can't help watching these three approaches as something not compatible, although I know it must be.

 

[mcastillo356]

Hi, PF

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

If there exists a constant $k$ bounding $|f(x)-P_{n}(x)|$ by $|x-a|^{n+1}$, we've got an order of approximation; this means an expression for how accurate the Taylor polynomial comes near to a function. Don't you think talking about accuracy is more precise than rapidlity?

 

[pbuk]

Don't you think talking about accuracy is more precise than rapidlity?

No, I think there must be some linguistic confusion. You are looking at two sides of the same coin and thinking it must be a different coin.

Because $f(x)=P_{n}(x)+O((x-a)^{n+1}) \iff |f(x)-P_{n}(x)|\leq{k|x-a|^{n+1}}$ we can talk about accuracy and rapidity interchangeably.

 

[mcastillo356]

It is a three sides coin, error, rapidity or accuracy, and order of approximation

 

[FactChecker]

It is a three sides coin, error, rapidity or accuracy, and order of approximation

You should not come away with the impression that the definition is sloppy. The casual descriptions here are a little deceptive. There are some formal definitions of the Big O notation that are quite specific. The details may differ among mathematicians, but each one is precise.

 

[mcastillo356]

Hi, PF, Big O notation seem it is understood, I'm going to publish the last obstacle, reef or rock where I stumble; the Theorem 13 at Calculus-A complete course 7th edition, R. Adams and C. Essex, Section 4.10, Taylor polynomials, It happens to be another thread I will untitle "Understanding T. 13, from the textbook Calculus, R.Adams and C. Essex, 7th edition, 4.10"

Love