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The statement of the theorem is that there exists a c in the open interval (a,b). Yet the very first line of the proof states that one can choose c to be either a or b. This is wrong. One must either relax the condition and let c be in the closed interval or prove the existence of such a c in the
1217:
Why does nobody write what the implications of this theorem are? In the Intro it says "But even when ƒ′ is not continuous, Darboux's theorem places a severe restriction on what it can be.". Exactly that is the question: What severe restrictions? This is one of the most important parts of the theorem
918:
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
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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
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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
915:
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
223:
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
923:
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).
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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).
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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.
946:
First usage of Ferma's theorem is wrong because you can use it on (a, b) and not . for example for function f(x) = x^2 on f has maximums on -2 and 2 (f(2) = f(-2) = 4) and derivation of f on neither -2 nor 2 is 0.
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After the sentence beginning "In particular, the derivative of the function" should come the sentence "This function is used in the construction of
Volterra's function.", with a link.
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To define the IVP, I think, is necessary to have a connected domain. In the cited work (Ciesielski) is for instance the real line. Otherwise we have also or a generic interval I.
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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.
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Nope. The first sentence of the proof shows the property of the function, which is then used to apply the extreme value theorem for
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901:, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.
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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.
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No, it was correct. The problem was the derivative symbol that was not clearly visible. I have added a thin space. --
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You are absolutely correct. The first mention of Fermat's theorem was not correct. I removed it. It's true that
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Note the location of the period at the end of the sentence in the display above. I changed it to this:
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916:'uncontrollably' out of respect of my lecturer I won't give out his examples here but they do exist.
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224:'uncontrollably' out of respect of my lecturer I won't give out his examples here but they do exist.
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on
Knowledge. If you would like to participate, please visit the project page, where you can join
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The statement of the theorem is wrong. Since the very beginning. In the original source,
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Last edited at 17:01, 3 November 2012 (UTC). Substituted at 01:58, 5 May 2016 (UTC)
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An amazingly extreme case of inattentiveness to what one is doing was this:
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should be merged with this one since the content has significant overlap.
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anyway hope i used the discussion feature correctly hear goes
322:, so Fermat's theorem tells us a local maximum cannot occur at
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In Proof: functions are not in general increasing/decreasing.
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Definition of intermediate value property of a function.
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636:{\displaystyle (\phi (a)-\phi (t))/(a-t)\leq 0}
944:== Wrong implementation of Ferma's theorem ==
891:The comment(s) below were originally left at
238:Hope this is in the right section this time
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813:{\displaystyle \phi '(a)=f'(a)-y: -->
528:{\displaystyle \phi (a)\geq \phi (t)}
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1327:. It does not extend the theorem. --
1218:and nobody seems to talk about it?
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38:It is of interest to the following
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1405:Mid-priority mathematics articles
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315:{\displaystyle \phi '(a)\neq 0}
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668:{\displaystyle a\leq t\leq b}
560:{\displaystyle a\leq t\leq b}
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158:Wrong from the very beginning
109:and see a list of open tasks.
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741:. But this contradicts that
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838:Lars Olsen
851:Mscdancer
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36:scale.
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206:McKay
1382:talk
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663:b
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631:0
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622:t
616:a
613:(
609:/
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599:t
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584:a
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479:a
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427:[
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330:a
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301:a
298:(
294:′
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42::
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