Talk:Taylor's theorem

message: The section Derivation for the mean value forms of the remainder exploits Cauchy's mean value theorem for the function
:t\in[x,a]\mapsto F(t) = f(t) + f'(t)(x-t) + \frac{f(t)}{2!}(x-t)^2 + \cdots + \frac{f^{(k)}(t)}{k!}(x-t)^k.
and an appropriate G.
To apply Cauchy's mean value theorem, F must be continuous on [x,a] and differentiable on (x,a). Consequently, one must require 1) that f^{(i)} exists and is continuous on [x,a] for all i=0,...k, AND 2) that f^{(k+1)} exists on (x,a).
On the other hand on section Explicit formulae for the remainder one reads:
Let f : R → R be k+1 times differentiable on the open interval and continuous on the closed interval between a and x.
It is not clear if continuous on the closed interval between a and x above refers to f or to each of the derivatives from i=0 to k+1. The first meaning would be definitely inconsistent with the given proof and 1)-2), the second one would be too much asking because the k+1 derivative is asked only to exist on (x,a).
Somebody that is aware of the correlations among the various sections of the article should fix this.
--R.ductor (talk) 08:47, 17 October 2013 (UTC)
replies: text: You're right, this is an error.  I have corrected it.  Sławomir Biały (talk) 11:23, 17 October 2013 (UTC)
thread_title:  Statement for the validity of Lagrange reminder inconsistent with the proof. 

message: While searching for a citation for the multi-dimensional theorem (see above discussion), I started to realise that all the sources I found that quoted the one dimensional theorem used stronger assumptions. I am talking about the very first boxed statement, under the heading "statement of the theorem". It only requires k derivatives at the point, but all the sources I've seen either require k continuous derivatives or k derivatives in an interval around the point. The exception is the Encyclopedia of Mathematics article currently cited in that section, but I don't think that counts as a reliable source: it is written by an expert, but I don't think it is peer-reviewed in the usual sense, and certainly doesn't have a proof. I tried chasing down the references quoted in that article, and managed to look at three of the six (the two Rudin books and the Apostol book) but they didn't verify this statement of the theorem. (Anyone trying to find citations might find it useful to know that this version is sometimes called "Peano's remainder". This might be worth adding to the article.) Quietbritishjim (talk) 00:28, 26 May 2012 (UTC)
replies: text: If there are no objections I will change this in a few days. Specifically I'll weaken the hypothesis from "k derivatives at a point x" to "k derivatives on an open interval about x". Quietbritishjim (talk) 12:55, 31 May 2012 (UTC)

text: The function only needs to be k times differentiable at the point.  I've added two more references to this effect.  Sławomir Biały (talk) 13:33, 31 May 2012 (UTC)

text: I've checked the first reference you added (Spivak's Calculus), and actually, Spivak fails to provide a proper definition of the function f he uses. While he only states that it must be n times differentiable, later he goes on to use L'Hopital's theorem, which requires that a function be differentiable over a whole neighborhood of the point. The same technique is used in the proof on the very same Wikipedia page, and therefore someone is yet to show a proof which does not require differentiability over a whole open interval. I think the statement needs to be changed to what Quietbritishjim proposed until someone comes up with a proof which does not require L'Hopital's theorem, or at least no differentiability about an open interval.Perrin4869 (talk) 08:54, 22 February 2013 (UTC)
replies: text: I think you (and the recent IP editor, if this is someone else) have misunderstood.  L'Hopital's rule is only needed to decrease the order from k to one.  At order one, the definition of the derivative can be used.  The proof in the article mentions this.  Note that differentiability up to order k at a point actually requires differentiability up to order k−1 in a neighborhood.  That's presumably the source of the confusion.  I have re-checked the references, and they do agree with the article as written.  Sławomir Biały (talk) 01:41, 1 March 2013 (UTC)
replies: text: "Note that differentiability up to order k at a point actually requires differentiability up to order k−1 in a neighborhood. That's presumably the source of the confusion.". Yes it is, I actually did not know this, and if this is true then I have no problem with the statement. I think the proof in the article should make mention of this to prevent confusion in the future. Do you have any citation to the claim that differentiability up to order k at a point requires differentiability up to order k-1 in a neighborhood? I'd very much like to see the proof of that. Oh, and by the way I was the recent IP editor, I had just forgotten to sign in before making the changes.Perrin4869 (talk) 02:25, 2 March 2013 (UTC)

text: Or actually, just thinking about it, it follows from the definition of the derivative, right? In order for a function to be differentiable, it must exist in a whole interval by definition, therefore if \frac{\mathrm{d}f}{\mathrm{d}x} is differentiable at a point a, then it must exist on a whole open interval, meaning that f is differentiable over an open interval. Am I correct?Perrin4869 (talk) 03:12, 2 March 2013 (UTC)
replies: text: Yes, that's right.  (With the usual allowances made for endpoints of the domain.)  Sławomir Biały (talk) 13:08, 2 March 2013 (UTC)
thread_title:  Main statement of one-dimensional theorem wrong? 

message: This section is in my opinion very misleading. If we have the estimate for the remainder term as in (*) (with respect to the original function) then the Taylor series does in fact converge to the original function, not just some analytic function. After the example of a Taylor series not converging to the original function is introduced, the sentence "Now the estimates for the remainder for the Taylor polynomials show that the Taylor series of f converges uniformly to the zero function on the whole real axis" is in my opinion wrong, because it's not the estimates of the remainder term that show us convergence of the series, but the simple fact that f^{(k)}(0)=0 \quad \forall k , so if we expand the series around zero, then T_f must be zero. If there were in fact uniform estimates of the remainder, would we not have that the series converges to the function? Consequently, i think the last bullet point is wrong.  — Preceding unsigned comment added by 217.162.129.176 (talk) 20:23, 19 March 2013 (UTC)
replies: text: See Taylor's theorem#Taylor's theorem in complex analysis. Sławomir Biały (talk) 11:58, 11 July 2016 (UTC)
thread_title:  Taylor's theorem and convergence of Taylor series 

message: There is a funny bug, I believe, at the page 1617 in science, that mentions Gregory.  According to the page about James Gregory (mathematician), Gregory was not yet born. Bdmy (talk) 08:47, 26 August 2013 (UTC)
thread_title:  1617 or 1671? 

message: This is used in the section on complex variables but is not defined anywhere in the article.  My guess would be that it means the ball of radius r centered at z, but that doesn't seem to make sense in the context.  In any case, the notation should be defined before it is used. Rick Norwood (talk) 17:24, 1 April 2015 (UTC)
thread_title:  B(z, r) 

message: Ladies Gentlemen,
Please permit me say that in second graph in Motivation variable k changes very speedy. 
With regards and friendship 
Georges Theodosiou  — Preceding unsigned comment added by 194.250.79.103 (talk) 09:43, 19 October 2015 (UTC)
thread_title:  Very speedy graph 

message: The letter  T is often used for the Taylor polynomials, instead of the most generic P. That is, the k-th order Taylor polynomial of f, centered at y, is denoted
:(0)\qquad T_k(x)=\sum_{j=0}^n\frac{f^{(j)}(y)}{j!}(x-y)^j , 
or even T_k(x,y,f) if we need to show the dependence from y and f. A complete notation has some advantages, for instance in the following easy but meaningful formulas: 
:(1)\qquad T_0(x,y,f)=f(y) 
:(2)\qquad T_k(x,x,f)=f(x) 
:(3)\qquad \partial_x T_k(x,y,f)=T_{k-1}(x,y,f') 
:(4)\qquad \partial_y T_k(x,y,f)= \frac{f^{(k+1)}(y)}{k!}(x-y)^k 
:(5)\qquad T_k(x,y,af+bg)=aT_k(x,y,f)+bT_k(x,y,g)  
 
For instance, formulas (1) (2) and (3) may give a reasonable inductive definition for (0); formula (4), already used in the paper for the proof of the mean value form of the remainder, also gives directly the integral form of the remainder, by a short plain application of the fundamental theorem of calculus, with no need of iterated integration  by parts, nor use of induction (assuming f of class C^{k+1} or just  f^{(k)} AC)
Proof: by formulas (2) and (4)
:R_k(x,a):=f(x)-T_k(x,a)=T_k(x,x)-T_k(x,a)=\int_a^x \partial_t T_k(x,t)dt=\int_a^x \frac{f^{(k+1)}(t)}{k!}(a-t)^kdt. 
In fact, it also gives another popular derivation of the Lagrange form of the remainder, directly from the Rolle's theorem, assuming f derivable k+1 times
Proof: Given x and a\neq x, we define a number r by the equality f(x)=T_k(x,a)+r(x-a)^{k+1}
By this choice of r, the function g(t):=T_n(x,t)+ r(x-t)^{n+1} has g(a)=f(x) but, it also has g(x)=f(x), therefore by the Rolle's theorem there is a number  c strictly between a and x with g'( c)=0, which in turn gives \displaystyle r=\frac{f^{k+1}(c)}{(k+1)!} and R_k(x,a)=\frac{f^{k+1}(c)}{(k+1)!}(x-a)^{k+1}.
Therefore, stating formula (4) once initially, may allow to prove the three forms in a more compact and unified way, and easier to remember. I'd prefer not to interfere with the current state of the article in it's main parts, but I'd be glad to have some feedback especially on the quite revolutionary idea of replacing P by T. I see that T is already used here and there in the article. pma 20:49, 20 November 2019 (UTC)
thread_title:  On notation, and other stuff

message: OK I think the following sentence could be problematic from a student's point of view:
"Instead of just matching one derivative of f(x) at x = a, this polynomial has the same first and second derivatives, as is evident upon differentiation."
I say this because even though the example is of a fitting to the exponential in the plots to the right, the statement should refer either to the example function or explicitly tie it into the exponential function as depicted in the figure. Experienced readers would know this, but for students, every statement should be accompanied by maximum clarification. Groovamos (talk) 04:37, 7 July 2021 (UTC)
thread_title:  Clarification of a particular statement for students' benefit 

message: For both Lagrange and Cauchy form the article says "for some real number c between a and x".
As explained in footnote 7 one has to be cautious if the interval is open or closed, so it would be better to explicitly
write a<c<x or a<=c<=x, respectively instead of "in between". Reading through reference (Apostol 1967, §7.7) which is cited as a source for this representation, it even seems that the interval has to be closed for Lagrange form and open for Cauchy form. Maybe someone can make this more clear. 92.213.82.9 (talk) 12:03, 30 August 2022 (UTC)
thread_title:  Explicit formulas for the remainder 

message: In the boxed theorem "Multivariate version of Taylor's theorem" the remainder sum should be for |alpha|=k, not k+1 (otherwise there are counterexamples) and if one wants to be faithful to the given reference [14] (in German: Königsberger Analysis 2, p. 64; a ref in English would be appreciated) one should assume in the assumptions that f is k+1 times differentiable, not just k times (and in a neighbourhood of the point, not just at the point). Note also that the remainder does not need to be in the form of the given sum: writing h(x)||x-a||^k is equivalent and easier to understand (no sum, no multi-index, ||x-a|| denotes the (any) norm of the vector x-a).
Actually, the reference [14] gives the stronger remainder forms that follow the boxed theorem.
But maybe instead of just making the corrections above one should present things differently. There is (see the French Wikipedia) a multivariate Taylor expansion theorem up to order k with a little o(||x-a||^k) remainder for maps that are only C^k (and even weaker assumptions work), called "Taylor Young", and they give a  reference in English. Taylor Young ought to appear in the boxed theorem. Then one could state below it the C^k+1 version of [14], with its precise remainders. Arnaud Chéritat (talk) 22:43, 3 November 2022 (UTC)
thread_title:  Multivariate Taylor theorem needs minor corrections in Aug 10 2022 version 

message: Should subscript 'C' be 'S' in the equation giving the Schlömilch form of the remainder? 111.220.172.117 (talk) 22:49, 11 January 2024 (UTC)
replies: text: yes, fixed, thanks. McKay (talk) 01:52, 15 January 2024 (UTC)
thread_title:  Schlömilch form of remainder