Talk:Darboux's theorem (analysis)

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Wrong from the very beginning

The statement of the theorem is wrong. Since the very beginning. In the original source, [1], it is correct. --Fioravante Patrone en (talk) 01:36, 11 February 2008 (UTC)

No, it was correct. The problem was the derivative symbol that was not clearly visible. I have added a thin space. --Fioravante Patrone en (talk) 10:12, 7 May 2008 (UTC)

Propose merge

The article Darboux function should be merged with this one since the content has significant overlap. McKay (talk) 08:53, 21 April 2009 (UTC)

In Proof: functions are not in general increasing/decreasing.

Hi, just a small point. Since derivatives are not necessarily continuous we can't say that functions in general are decreasing or increasing in a neighbourhood of a point regardless of the value of the derivative there. You can create examples easily enough by considering functions that oscillate 'uncontrollably' out of respect of my lecturer I won't give out his examples here but they do exist.

What can be said in the context of your proof of Darboux theorem is that in a right neighbourhood of a there exists a y such that f(a)<f(y)(since as the function was differentiable the limit f(a)- f(x)/x-a existed thus was positive in the right neighbourhood) thus the max is to the right of a and similar for b.

just to labour the point saying the function increases in a right-neighbourhood of a says that if x, y belong to that neighbourhood and x lies between a and y then f(x)<f(y). However all that is implied by the derivative taking a positive value at a is that f(a)<f(x) and f(a)<f(y).

Hope this is in the right section this time

regards 122.110.191.137 (talk) 16:28, 28 October 2009 (UTC) CWC

Fermat's Theorem

The proof of the theorem mention's Fermat's theorem but I don't think it is quite Fermat's theorem although it is related. Pratyush Sarkar (talk) 02:06, 23 July 2013 (UTC)

You are absolutely correct. The first mention of Fermat's theorem was not correct. I removed it. It's true that ${\displaystyle \phi '(a)\neq 0}$, so Fermat's theorem tells us a local maximum cannot occur at ${\displaystyle a}$. But we are trying to argue that the absolute maximum on ${\displaystyle [a,b]}$ cannot occur at ${\displaystyle a}$.
Here is the reason the absolute maximum of ${\displaystyle \phi }$ on ${\displaystyle [a,b]}$ cannot occur at ${\displaystyle a}$. If the absolute maximum did occur at ${\displaystyle a}$, then ${\displaystyle \phi (a)\geq \phi (t)}$ for all ${\displaystyle a\leq t\leq b}$. This implies ${\displaystyle (\phi (a)-\phi (t))/(a-t)\leq 0}$ for all ${\displaystyle a\leq t\leq b}$, and letting ${\displaystyle t\rightarrow a}$ gives ${\displaystyle \phi '(a)\leq 0}$. But this contradicts that ${\displaystyle \phi '(a)=f'(a)-y>y-y=0}$. LoonetteTheClown (talk) 15:10, 4 September 2016 (UTC)

Lars Olsen

The article says the proof based on Fermat's theorem and its corollary is due to Lars Olsen. But I have read exactly the same proof from much older text. — Preceding unsigned comment added by Mscdancer (talkcontribs) 13:21, 24 November 2015 (UTC)

The article currently says "Another proof based solely on the mean value theorem and the intermediate value theorem is due to Lars Olsen," and cites Olsen, Lars: A New Proof of Darboux's Theorem, Vol. 111, No. 8 (Oct., 2004) (pp. 713–715), The American Mathematical Monthly. The proof being referred to is not due to Lars Olsen. Despite the title "A New Proof of Darboux's Theorem," Olsen's proof is not new. The proof appears, for example, in Mathematical Analysis (2e) by Tom M. Apostol. See Theorem 5.16 of that book. LoonetteTheClown (talk) 13:33, 4 September 2016 (UTC)

Assessment comment

The comment(s) below were originally left at Talk:Darboux's theorem (analysis)/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Last edited at 17:01, 3 November 2012 (UTC). Substituted at 01:58, 5 May 2016 (UTC)

imbecilic punctuation

An amazingly extreme case of inattentiveness to what one is doing was this:

${\displaystyle x\mapsto {\begin{cases}\sin(1/x)&{\text{for }}x\neq 0\\0&{\text{for }}x=0\end{cases}}}$.

Note the location of the period at the end of the sentence in the display above. I changed it to this:

${\displaystyle x\mapsto {\begin{cases}\sin(1/x)&{\text{for }}x\neq 0,\\0&{\text{for }}x=0.\end{cases}}}$