Talk:Piecewise property

Latest comment: 1 month ago by Maddy from Celeste in topic Requested move 26 August 2024

Origin

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A disambigation was requested at Talk:Piecewise_function#Requested_move_20_July_2024. I split the part here that did not apply to the new title. 142.113.140.146 (talk) 00:55, 25 August 2024 (UTC)Reply

Requested move 26 August 2024

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The following is a closed discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review after discussing it on the closer's talk page. No further edits should be made to this discussion.

The result of the move request was: no consensus. Clarity did not develop. (closed by non-admin page mover) -- Maddy from Celeste (WAVEDASH) 14:15, 3 October 2024 (UTC)Reply


Piecewise propertyPiecewise – The "piecewise" isn’t a property in the way, say, the continuity is a property of a function. The same point has been made by other editors at Talk:Piecewise_function#Requested_move_20_July_2024. Taku (talk) 08:08, 26 August 2024 (UTC) — Relisting. Steel1943 (talk) 21:20, 17 September 2024 (UTC) — Relisting. BD2412 T 01:01, 25 September 2024 (UTC)Reply

Relisting comment: Relist one more time to see if clarity develops BD2412 T 01:01, 25 September 2024 (UTC)Reply
The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Definition

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(refactored from Talk:Piecewise_property#Requested_move_26_August_2024)

Is piecewise [continuous differentiable] [really] a property of the function itself? The absolute value piecewise-defined function is. But counterexample:   with domain restriction   is non-differentiable at   and is defined using 1 subfunction. Redefine it with 2 identical subfunctions and it becomes piecewise differentiable. These are contradictory properties as "the function itself" is the same under equality of   pairs. 142.113.140.146 (talk) 10:51, 26 August 2024 (UTC) 174.89.12.36 (talk) 23:44, 11 September 2024 (UTC)Reply
It is always piecewise differentiable, doesn't matter how you define it. Tercer (talk) 14:09, 26 August 2024 (UTC)Reply
So you're saying it's iff it admits a definition that is. The article should say that. I added it. 142.113.140.146 (talk) 18:20, 26 August 2024 (UTC) 174.89.12.36 (talk) 23:44, 11 September 2024 (UTC)Reply
I don't really see where that College Board source you cited asserts a different meaning of piecewise differentiability. SilverLocust 💬 22:57, 4 September 2024 (UTC)Reply

@SilverLocust: The College Board source's Theorem 2 requires the definition to be examined for   and  . This is not a different meaning but the article's original definition. The next 2 sources contain the generalization in this thread. The Nikolsky source does not require a definition (see no {), and supports the new claim doesn't matter how you define it. 174.89.12.36 (talk) 23:54, 11 September 2024 (UTC)Reply

It doesn't even mention piecewise differentiability anywhere, just differentiability of a piecewise-defined function at a point. SilverLocust 💬 00:26, 12 September 2024 (UTC)Reply
Idk how to fix this, so I put up a {{bsn}}. I prefer Tercer's generalization, but the rest of the article uses the version that looks at the subfunctions. 174.89.12.36 (talk) 01:44, 12 September 2024 (UTC)Reply