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

In summary: I mean my poor english.In summary, the theorem states that the distance between the graphs of the Taylor and MacLaurin polynomials approaches infinity as the graph approaches the original function.
  • #1
mcastillo356
Gold Member
560
267
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).
IMG_20220514_151516.jpg
 
Physics news on Phys.org
  • #2
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.:sleep:
 
  • #3
Hi PF, working on it. I think must make an effort I haven't made still.
Greetings
 
  • #4
This follows from any of the explicit expressions for the remainder. For example, if [tex]f(x)
= f(a) + (x - a)f'(a) + \dots + \frac{(x - a)^n}{n!}f^{(n)}(a) + R_n(x)[/tex] then by applying the mean value theorem to [tex]F(t) = f(t) + (x - t)f'(t) + \dots + \frac{(x - t)^n}{n!} f^{(n)}(t)[/tex] it follows that there exists [itex]\xi[/itex] between [itex]a[/itex] and [itex]x[/itex] such that [tex]
\begin{split}
R_n(x) &= F(x) - F(a) \\
&= (x- a)F'(\xi) \\
&=
(x - a) \frac{(x - \xi)^{n}}{n!}f^{(n+1)}(\xi).\end{split}[/tex]
 
  • Like
  • Informative
Likes vanhees71, WWGD and mcastillo356
  • #5
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.
 
Last edited:
  • Like
Likes WWGD and pbuk
  • #6
mcastillo356 said:
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.
 
  • Sad
Likes mcastillo356
  • #7
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:smile:
 
  • #8
Hi, PF
pasmith said:
This follows from any of the explicit expressions for the remainder. For example, if [tex]f(x)
= f(a) + (x - a)f'(a) + \dots + \frac{(x - a)^n}{n!}f^{(n)}(a) + R_n(x)[/tex] then by applying the mean value theorem to [tex]F(t) = f(t) + (x - t)f'(t) + \dots + \frac{(x - t)^n}{n!} f^{(n)}(t)[/tex] it follows that there exists [itex]\xi[/itex] between [itex]a[/itex] and [itex]x[/itex] such that [tex]
\begin{split}
R_n(x) &= F(x) - F(a) \\
&= (x- a)F'(\xi) \\
&=
(x - a) \frac{(x - \xi)^{n}}{n!}f^{(n+1)}(\xi).\end{split}[/tex]
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
 
  • #9
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!
 
  • #10
mcastillo356 said:
"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##.

mcastillo356 said:
(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}##
 
  • #11
mcastillo356 said:
(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)|##
 
  • #12
mcastillo356 said:
(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##

mcastillo356 said:
##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))##
 
  • #13
Still, @mcastillo356 , notice Sinx is somewhat of a special case, as ## |sinx|\leq 1##
 
  • Informative
Likes mcastillo356
  • #14
WWGD said:
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.
 
Last edited:
  • #15
mcastillo356 said:
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.
 
Last edited:
  • #16
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?
 
  • #17
mcastillo356 said:
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.
 
  • Like
  • Informative
Likes FactChecker and mcastillo356
  • #18
It is a three sides coin:wink:, error, rapidity or accuracy, and order of approximation
 
  • #19
mcastillo356 said:
It is a three sides coin:wink:, 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.
 
  • Like
  • Love
Likes mcastillo356 and pbuk
  • #20
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
 

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

1. What is Big-O notation and why is it important in computer science?

Big-O notation is a mathematical notation used to describe the asymptotic behavior or growth rate of a function. It is commonly used in computer science to analyze the time and space complexity of algorithms. It helps in comparing the efficiency of different algorithms and choosing the most efficient one for a given problem.

2. How does Big-O relate to the Taylor graph and the function f(x)?

Big-O notation is used to measure the rate at which the Taylor graph of a function approaches the graph of the function itself. This helps in understanding the behavior of the function and its efficiency in terms of time and space complexity.

3. Can you provide an example of how Big-O is calculated for a function?

Let's take the function f(x) = 2x + 5. The Taylor graph of this function would be a straight line with a slope of 2 and an intercept of 5. The Big-O notation for this function would be O(x), as the Taylor graph approaches the graph of f(x) at a linear rate.

4. How can Big-O notation be used to compare different algorithms?

Big-O notation allows us to compare the time and space complexity of different algorithms. The algorithm with a lower Big-O complexity would be considered more efficient as it would have a faster runtime and use less memory compared to an algorithm with a higher Big-O complexity.

5. Are there any limitations to using Big-O notation?

Big-O notation only gives an asymptotic upper bound on the growth rate of a function, and it does not take into account the constant factors or lower order terms. This means that two algorithms with the same Big-O complexity may have different actual runtimes. Additionally, Big-O notation does not consider the worst-case, average-case, or best-case scenarios of an algorithm, which may affect its efficiency in different situations.

Similar threads

Replies
3
Views
1K
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
893
Replies
24
Views
2K
Replies
9
Views
5K
Replies
5
Views
1K
  • Calculus
Replies
1
Views
1K
  • STEM Educators and Teaching
Replies
4
Views
1K
  • Topology and Analysis
Replies
10
Views
2K
Back
Top