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Talk:Piecewise property

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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]