# Talk:Taylor series

I just looked at this article with the edge browser and the x^4 and x^5 powers looked weird in the first version of the Maclaurin series for the exponential function, the one with factorials. They look fine in chrome! must be a bug of some sort? — Preceding unsigned comment added by 2a01:388:2ec:150::1:12 (talk) 13:25, 28 September 2021 (UTC)

Must be -- probably you should use a better browser than Edge. --JBL (talk) 13:53, 28 September 2021 (UTC)

Please fix the "Examples" section to make sense. For example, 
1st example is about the "MacLaurin series". I'm here to learn about the Taylor series. I see MacLaurin mentioned in 2nd paragraph but I wasn't sure what "about zero" means.  Maybe if the example section had a Taylor series example... 
4th line says  "so the Taylor series..." What does it mean by "so". How does that so obviously follow from the MacLaurin example? I can't see the connection. 
6th line says "By integrating the above Maclaurin series"   What? You just called it a Taylor series. Why are you integrating it? That's not even what I get when I integrate it.
And so on.
I would like to see some examples in the "examples" section.
Ywaz (talk) 00:43, 14 October 2021 (UTC)

There is no possible way to write an article that will make it understandable to a person who has decided not to try to read it. --JBL (talk) 01:14, 14 October 2021 (UTC)
No, I agree with Ywaz. This is yet another appallingly written mathematical article which seems to assume the reader already knows what the author is talking about. I likewise got no further than the "Examples" section before I was totally baffled.
212.159.76.165 (talk) 16:33, 14 June 2022 (UTC)
I have fixed the grammar of section § Examples (misuse of "for" instead of "to", and comma after "so"). I have also added a short explanation to "so". For the remainder of Ywaz's complaints, nothing more can be done: "about 0" is not in the article, "Maclaurin series" is defined twice, in the lead and in section § Definition, and can thus be supposed to be known. About "appallingly written", I would be happy if you could propose a better way to write this article. But, as for every technical article, a minimal background is required for understand it. Here, nobody can understand the subject without having learnt first what is a series and a derivative. D.Lazard (talk) 17:35, 14 June 2022 (UTC)
I agree the examples section does not flow very well and would be helped by more steps to show how it works for people trying to learn this, rather than just stating the results.  Another problem with this section is that the part that starts with "By integrating the above Maclaurin series" seems to be not quite right.  It is the integral of -1 times that series.  (You don't get -x by integrating 1).   It would also be helped by referring to the series for (it should be) -1/(x-1) or 1/(x-1) by a name or other identifier (like a parenthesized formula number as is often done) so the reader doesn't think it is referring somehow to the immediately preceding series. Skaphan (talk) 18:18, 10 February 2023 (UTC)
I am also having trouble understanding this section.  Right at the beginning, it says:

The Taylor series of any polynomial is the polynomial itself.
The Maclaurin series of ⁠1/1 − x⁠ is the geometric series

  
    
      
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    {\displaystyle 1+x+x^{2}+x^{3}+\cdots .}
  

So, by substituting x for 1 − x, the Taylor series of ⁠1/x⁠ at a = 1 is

  
    
      
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    {\displaystyle 1-(x-1)+(x-1)^{2}-(x-1)^{3}+\cdots .}
  

I am having trouble understanding everything after "So, by substituting..." Why?  Why are we substituting x for 1-x?  How does this follow from the first Maclaurin series?  It seems to have jumped past some explanation of what is being discussed.
Wikinetman (talk) 18:46, 16 February 2024 (UTC)

User:JayBeeEll thinks that the lead sentence should define Taylor series as "infinite sums" rather than power series so as to avoid jargon.
However:

I provided a reference stating that Taylor series are power series, whereas no reference indicates that Taylor series are "infinite sums"
An "infinite sum" has literally no meaning, neither mathematical nor practical (because it is of course impossible to add up infinitely many numbers or functions). Maybe it could serve as a introductory pedagogical term, but then I think the correct term would have to be included.
The phrase "infinite sum" is ambiguous as it could mean many things, like a numerical series, a series of functions...
Thus I would like to revert the last edit. --L'âne onyme (talk) 20:47, 27 October 2021 (UTC)

The fact that you do not accept universally understood terminology is perhaps an interesting personal foible, but it is not a basis for editing Wikipedia articles.  Indeed, there are many kinds of infinite series -- just as there are many kinds of power series, many kinds of polynomials, many kinds of anything.  It is my belief that essentially everyone learns of Taylor series as the first kind of (infinite) power series they meet -- ergo saying "Taylor series is a kind of power series" is defining a more common and familiar term in terms of a less common one.  This is in comparison to the current text, which introduces a potentially new idea in terms of more widely familiar ideas. --JBL (talk) 21:19, 27 October 2021 (UTC)
Note, the context is Talk:Complex analysis#Difference between "Series" and "Sum of a series" –jacobolus (t) 02:39, 28 October 2021 (UTC)
Although the formulation "a Taylor series is a series, ..." is a sort of pleonasm, it is ambiguous, as a series has many meanings (readers may have skip the first phrase "in mathematics"). So, "is an infinite sum" (with a link to Series (mathematics)) is better, as it recalls to non-expert readers what is a mathematical series. Your argument that the phrase has "literally no meaning" is a fallacy, as all definitions given in textbook use it, or, such as series (mathematics) does, use a similarformulation. D.Lazard (talk) 08:48, 28 October 2021 (UTC)

At this time, the "examples" section says: 
"So, by substituting x for 1 − x, the Taylor series of ⁠1/x⁠ at a = 1 is

  
    
      
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    {\displaystyle 1-(x-1)+(x-1)^{2}-(x-1)^{3}+\cdots .}
  

By integrating the above Maclaurin series, we find the Maclaurin series of ln(1  − x), where..." 
The example identifies a = 1 for the Taylor series, but then calls it a Maclaurin series, which should require a = 0. In other words, "by integrating the above Maclaurin series" is not correct, the equation above is a Taylor series, not a Maclaurin series. Therefore, I suggest this be edited to: 
"By integrating the above Taylor series, we find the Maclaurin series of ln(1  − x), where..."
I humbly present this concern recognizing I could be mistaken!
— Preceding unsigned comment added by Caireau (talk • contribs) 10:50, 16 July 2024 (UTC)

It's a bit tricky perhaps because the variable name is being re-used. To clarify by using different variable names, the claim is that the "Maclaurin series" for the function ⁠
  
    
      
        
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    {\displaystyle \textstyle u\mapsto \int {du/(1-u)}=\ln(1-u)}
  
⁠ centered at ⁠
  
    
      
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    {\displaystyle u=0}
  
⁠ is the opposite of the "Taylor series" at ⁠
  
    
      
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    {\displaystyle x=1}
  
⁠ for the function ⁠
  
    
      
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    {\displaystyle x\mapsto \ln x}
  
⁠, under the change of variables ⁠
  
    
      
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    {\displaystyle x=1-u}
  
⁠. Aside: while introductory calculus textbooks nowadays tend to be very pedantic about a distinction between "Taylor series" and "Maclaurin series" in actual mathematical literature these are routinely conflated and given either name at the author's whim, and it's seldom if ever confusing in context. –jacobolus (t) 13:00, 16 July 2024 (UTC)
I follow. Thanks for that clarification.
I can see how the difference between Maclaurin and Taylor series can be pedantic since there's nothing special about a=0. The prose of the explanation seems to have tripped me up in this case. Caireau (talk) 10:45, 17 July 2024 (UTC)
I think the section is definitely at least mildly confusing and could probably be copyedited/rewritten for clarity, if anyone wants to take a crack at it. –jacobolus (t) 15:41, 17 July 2024 (UTC)

The section "List of Maclaurin series of some common functions" says the Maclaurin for the Exponential function converges for all values of X.  I disagree.  When I first took calculus in the AP match in high school, I studied the Maclaurin series and learned it converges for values of X close to 0.  In the Taylor Series, the value of X needs to be close to A.  From what I understand, when computing the Exponential function, one takes the integer portion of the exponent and calculate the value by the Power function, or use an array to get the value, then plugged the decimal portion of the exponent into the Maclaurin series, and multiply the two results.  
I set up an Access VBA stored procedure that calculates the value of 88.21389 of the Exponential function using Maclaurin series, subtract it from the value plugged into the EXP function, and the difference is "1.54269E+37". Dkf12 (talk) 18:30, 26 September 2024 (UTC)

A general Taylor series only converges within a disk around its center, and in general the radius of that disk is set by wherever the closest singularity is. But the series for the exponential function in particular converges in the entire complex plane. I'm not sure what's happening with your VBA script, but (a) in general I would not recommend implementing numerical routines in VBA, and (b) just because a series converges doesn't mean it is an effective computation tool in practice. –jacobolus (t) 20:11, 26 September 2024 (UTC)
My VBA script is accurate as I described.  Please refer to the website I found on Maclaurin series:

   https://brilliant.org/wiki/maclaurin-series/

It says quite clearly, "A Maclaurin series is a power series that allows one to calculate an approximation of a function f(x) for input values close to zero..."
As I noted, when I took calculus for AP math in high school, I learned the value of x of f(x) in a Maclaurin series needs to be close to 0.  That makes perfect sense.  To find values greater or less than 0, one needs to calculate the power or root of the integral portion, use Maclaurin for the decimal portion, and multiply the results.  — Preceding unsigned comment added by Dkf12 (talk • contribs) 18:33, 6 March 2025 (UTC) 

"close to zero" does not excludes "far to zero". The Maclaurin series of the exponential function gives approximations of the function near to zero, but, for this specific function, one gets approximations for every ⁠
  
    
      
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    {\displaystyle x}
  
⁠, although they need more terms lor large ⁠
  
    
      
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    {\displaystyle x}
  
⁠. D.Lazard (talk) 19:59, 6 March 2025 (UTC)
The exponential function is an entire function meaning it can be expressed as a sum of a single power series, in this case 
  
    
      
        
          
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    {\displaystyle \textstyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}}
  
 for all 
  
    
      
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    {\displaystyle x\in \mathbb {R} }
  
. See also radius of convergence of a power series, which you can calculate as

  
    
      
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    {\displaystyle R=\left(\limsup _{n\to \infty }{\sqrt[{n}]{|c_{n}|}}\right)^{-1}=\left(\limsup _{n\to \infty }{\sqrt[{n}]{\frac {1}{n!}}}\right)^{-1}=+\infty }
  

I don't know how accurate your script is but when I plug the value 88.21389 into a simple C++ script (with floats and no optimization for rounding errors, just a simple sum from the first to the last) I get within 0.0001% of the correct result (2.04553e+38) after just 140 terms. So either you have a very big rounding error or some other problem (I am not even getting your difference for any partial sum). MartinVitVavrik (talk) 18:20, 7 March 2025 (UTC)
Try plugging different values into the formula for Mclaurin.  I find that the result is far different from the value of EXP(X) when I plugged in 57 for X.  Values less than 57 do work out.  — Preceding unsigned comment added by Dkf12 (talk • contribs) 22:19, 9 March 2025 (UTC) 

That's a very good indication of exactly where your implementation of large integer arithmetic is being swamped by overflow or other rounding errors.  The convergence of the power series for the exponential function is a very straightforward application of the ratio test, and this fact is covered in every standard calculus textbook (e.g., it's Example 6.16 here).  --JBL (talk) 22:52, 9 March 2025 (UTC)