Talk:Taylor's Theorem

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I changed the formatting on the 3 block of LaTeX under Integral Version since it wasn't parsing. However, I would be curious to know if there is a way to align equals signs (which is what I assume was intended) in wiki formatting. If anyone knows, tell me so I don't have to go through 20 different wikipedia manual pages to find out. Thanks. --cynic 03:36, 18 September 2008 (UTC)

My usual technique is to use a table. See my user page or some of the posts I've put up for how I do this. It's not ideal but it works for me. --prime mover (talk) 05:15, 18 September 2008 (UTC)

You could also use:
$ \begin{matrix} =& 2+2\\ =&4 \end{matrix} $
See the code for this page. It just uses matricies.--Joe 11:41, 18 September 2008 (UTC)

Trouble with that is that it centralises the RHS and it usually looks better to left justify it. What I've done is set myself up a template for a proof which has a right justified LHS, centre justified bit in the middle (usually a relation or something) and a left justified RHS. If there is no LHS (as is usual in the way I present my proofs and extended equations), that column can be deleted. --prime mover (talk) 21:23, 18 September 2008 (UTC)

Thanks for cleaning this up guys! --Jehan60188 20:44, 23 September 2008 (UTC)

Questionable proof

This page seems to just claim a simplified version of taylor's theorem and then prove it by quoting taylor's theorem more specifically... I want to insert a "questionable" marker. --Brendan Wallace (talk) 01:09, 14 October 2012 (UTC)

I'm sorry, I don't understand. I can count three proofs. Are they all complete rubbish? If so then I need to take a view on my competence, and perhaps resign my position. --prime mover (talk) 06:09, 14 October 2012 (UTC)
I get the impression that you missed the link to Taylor's Theorem/One Variable (it's the header of the section you mistook for a proof). --Lord_Farin (talk) 07:32, 15 October 2012 (UTC)
Yep it's ok; though it's not that clear that you have to follow the link for the proof, since there's just one theorem on the page at the moment (I had to read Lord Farin's comment before I realised the proof was there).
Unrelatedly; (and possibly irrelevantly, depending on what's meant by the WIP flag): as it stands the theorem sounds like a Stone-Weierstrass-type claim. I suggest it should be changed to something else entirely, rather than specifying what 'approximated by polynomials' means, since it'll generally be assumed that this means some kind of convergence of the remainder term. --Linus44 (talk) 09:15, 15 October 2012 (UTC)
Quite possibly. The initial page was shamelessly cut and pasted from Wikipedia in the embryonic days of ProofWiki by someone who has not been heard from recently. Whether that person had any plans to make this rigorous is anyone's guess. The "one variable" version was developed subsequently and somewhat more rigorously. Refactoring welcome. --prime mover (talk) 10:24, 15 October 2012 (UTC)
The one-variable case is up for a review as well; I recall the definition of $R_n$ as a complicated integral, but here it is just taken to be some (scaled) value of $f^{(n+1)}$, while we didn't assume $f$ was $n+1$ times differentiable. If we do, the theorem statement enters a non-terminating recurrence and we need to assume $f$ smooth. --Lord_Farin (talk) 10:31, 15 October 2012 (UTC)
It did actually say $n$ times differentiable - so I've fixed it to say $n+1$ times. Does that work? --prime mover (talk) 12:00, 15 October 2012 (UTC)

I think it's fine now. However it's a bit hard to tell since my most authoritative source on this matter deals with $\R^m$ in place of $\R$. A bit that still needs fixing is the statement that $x \le \eta \xi$, since it may well be that $\xi < x$. --Lord_Farin (talk) 12:22, 15 October 2012 (UTC)

Yes, good point. We really do need a notation for "between", which in this context would mean: "$x \le \eta \le \xi$ if $x \le \xi$ but $\xi \le \eta \le x$ if $\xi \le x$" because this came up the other day in a different context. --prime mover (talk) 12:53, 15 October 2012 (UTC)
I suppose that $\eta \in \operatorname{ch} \left\{{x, \xi}\right\}$ (where $\operatorname{ch}$ denotes convex hull) is a bit overcomplicated in this situation? :) --Lord_Farin (talk) 12:57, 15 October 2012 (UTC)
There's a quite general proof that follows from the one variable case: if $f : \R^n \to \R^m$, $x \in \R^n$, we define $\phi : [0,1] \to \R^m$, $\phi(t) = f(x + t(y-x))$, then apply the one-variable version to $\phi$.
In particular this shows that $\xi \in [x,y] = \{x + t(y - x) : t \in [0,1]\} = \operatorname{ch}\{x,y\}$. The same goes for any Banach space. --Linus44 (talk) 16:13, 15 October 2012 (UTC)
Yes what you say is quite true, but this statement and proof is accessible to people whose education stopped at undergrad. We do of course need to put the general version up, but we mush not lose the basic analysis result which is familiar to school-level education. --prime mover (talk) 18:54, 15 October 2012 (UTC)
Yep. We'd need the real function version anyway to be able to do the proof this way. --Linus44 (talk) 12:40, 16 October 2012 (UTC)
Got an exposition of the theorem from Larson, but it might be a good idea to concoct a "betweenness" notation to make it prettier. I created $\mathsf{B}abc$ when I put up the Tarski axioms if you want to do something with that --GFauxPas (talk) 05:48, 16 October 2012 (UTC)

Remainder terms

Ok, with the notation of Leibniz's Rule/Real Valued Functions, the theorem for $f : \R^n \to \R\text{ or } \C$:

$\displaystyle (1) \qquad f(x) = \sum_{|\alpha| \leq p} \frac{(x - x_0)^\alpha}{\alpha!} \partial^\alpha f(x_0) + (1 + p) \sum_{|\alpha| = p + 1} \frac{(x - x_0)^\alpha}{\alpha!} \int_0^1 (1-t)^p \partial^\alpha f(x_0 + t(x-x_0))\ dt$

with integral remainder.

The problem is there's as many expressions as you fancy for the remainder term. We can have integral remainder and the kind discussed above without proving everything twice as follows. First prove the statement $(1)$ above. Then we have what wikipedia calls the first mean value theorem for integration. Applying this to the remainder term; we get the Lagrange form of the remainder (again, wikipedia terminology; never heard it actually used). --Linus44 (talk) 12:40, 16 October 2012 (UTC)