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$f$ being any real valued function, $k1, k2$ being two positive scalars. The scenario is:

  • "do something" if: $$x-f(x)<k_1 ~~~~s.t.~~~|x|>k_2$$

Can I rewrite it as:

  • "do something" if $$x-f(x)<k_1~~~AND~~~ |x|>k_2$$ $$\implies \frac{x-f(x)}{|x|}<\frac{k_1}{k_2}$$
ryang
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lifezbeautiful
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    'do something if: $x-f(x)<k_1 ~~s.t.~|x|>k_2$' is not valid. – geetha290krm Feb 16 '23 at 06:12
  • what do you mean by 'not valid'? – lifezbeautiful Feb 16 '23 at 06:14
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    It does not make sense. – geetha290krm Feb 16 '23 at 06:17
  • Apologies for lack of formal language, "do something" can be: "compute some new function g(x) for x"; Does it make sense now? – lifezbeautiful Feb 16 '23 at 06:19
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    The usage of "s.t." in the first scenario sounds strange. What you probably mean is "$x-f(x)<k_1$, $\forall x\ s.t.\ |x|>k_2$" or "$|x|>k_2\implies x-f(x)<k_1$". – A.Γ. Feb 16 '23 at 06:21
  • @A.Γ. Yes, you are right, this is infact \forall; In the original paper they used s.t. so I retained it in the question but it is confusing. I edited the question. Does it make sense now? – lifezbeautiful Feb 16 '23 at 06:32
  • What you mean is $\forall x (|x|>k_2 \Rightarrow x-f(x)<k_1)$. Writing $\forall |x|$ is confusing and lazy. – AlvinL Feb 16 '23 at 06:36
  • @AlvinL, Yes, I think that is what I meant. I followed the notation in equation 1 of this well-cited paper (https://arxiv.org/abs/1706.05394). I can attach a screenshot of the original equation in the question if that makes it clearer. The original equation does not sound confusing at all though. Maybe I messed something up when I simplified it for asking here :) – lifezbeautiful Feb 16 '23 at 06:41
  • @AlvinL Yes, exactly, so does it mean that I can treat it as AND and combined both parts (both parts implies the part before and after s.t.)? – lifezbeautiful Feb 16 '23 at 06:51
  • I strongly encourage you to use fewer formal symbols and more English words when you write mathematics. The two statements in your post are very confusing. In the first one there is probably a missing quantifier, and in the second one it looks like you used $\implies$ but really meant "therefore", and then tried to cram all these statements inside the condition for the "if". Things would be a lot more understandable if you had split them into meaningful English sentences instead of trying to make things compact with formal symbols but without explaining the purpose of the inequations. – Stef Feb 16 '23 at 16:43
  • Deleted my answer, because it contains errors which have been addressed in the other answers. Thx to @whoisit for pointing to my mistake. – AlvinL Feb 16 '23 at 17:00
  • @Stef Sounds like you are summarising my answer below. – ryang Feb 16 '23 at 17:35
  • @ryang to be honest I did not understand your answer, because of the D Q P R – Stef Feb 16 '23 at 20:40
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    @Stef The OP started with a theorem <"Do something" if condition Q is satisfied>; they noticed that <condition Q implies R >; they then invalidly concluded that < D if (R is satisfied)>. $\quad$ If, however, they had noticed that < P implies condition Q >, then it would have been valid to conclude that < D if (P is satisfied)>. – ryang Feb 16 '23 at 21:03

2 Answers2

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  • "do something" if $$x-f(x)<k_1 ~~~~s.t.~~~|x|>k_2$$

Perhaps the author means this:

  • "do something" if $$|x|>k_2\implies x-f(x)<k_1.$$

Can I rewrite the scenario as:

  • "do something" if $$x-f(x)<k_1~~~AND~~~ |x|>k_2$$ $$\implies \frac{x-f(x)}{|x|}<\frac{k_1}{k_2}$$

Given that $``D \text{ if } Q"$ and that $\color{cyan}P⟹Q$ and $Q⟹\color{violet}R,$ then:

  1. these are meaningful but invalid
    • $D\text{ if }\color{violet}R\color\red{\quad\quad\longleftarrow\text{You meant this.}}$
    • $D\text{ if }(Q⟹\color{violet}R)\color\red{\quad\quad\longleftarrow\text{You wrote this; it isn't equivalent to the previous line.}}$
    • $D\text{ if }(Q\text{ implies }\color{violet}R)\color\red{\quad\quad\longleftarrow\text{You wrote this.}}$
  2. these are valid
    • $D \text{ if }\color{cyan}P$
    • $(D \text{ if }Q);\text{ therefore }(D \text{ if }\color{cyan}P)$
    • $(D \text{ if }Q),\text{ which implies }(D \text{ if }\color{cyan}P)$
  3. this is also valid but conveys the least information as it does not claim that $(D \text{ if }Q)$ is actually true
    • $(D \text{ if }Q)\text{ implies }(D \text{ if }\color{cyan}P).$
ryang
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1

"Such that" is only used paired with an "There exists". The condition should be

"do something" if: $$\text{there exists}~~x~~\text{ such that}~~~ x-f(x)<k_1 ~~~~\text{AND}~~~|x|>k_2$$

Daron
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  • @ryang: The second one of those works; the first one doesn't really, but a similar example that does could probably be constructed :) – psmears Feb 16 '23 at 15:22
  • @ryang: Yes, but it's so contrived that it just sounds wrong. A less contrived example would probably sound a lot better :) – psmears Feb 16 '23 at 15:25
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    @ryang In that case you cannot replace "such that" with "and". After all, "If is and it is negative. . . " is nonsense. – Daron Feb 16 '23 at 16:13
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    I disagree that "such that" is only used paired with "there exists". For instance, the OP's $x-f(x)<k_1 ~~s.t.~|x|>k_2$ would make sense with an added "for all $x$" before the "such that". – Stef Feb 16 '23 at 16:38
  • @Daron Not following, nor getting what your fragmented, incomplete, indeed, nonsensical example is with reference to. $\quad$ I'll give a simpler example: does "If someone is cold such that they are not freezing, then they are fine" sound existentially- or universally-quantified? $\quad$ More to the point: one common and legitimate way to read ∀x(Px⇒Qx) is "for all x such that Px is true, Qx is true". – ryang Feb 16 '23 at 17:32