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Jan 10, 2023 at 12:00 history tweeted twitter.com/StackMma/status/1612781306429034499
Jan 5, 2023 at 3:32 answer added ubpdqn timeline score: 2
Jan 3, 2023 at 20:58 history edited David G. Stork CC BY-SA 4.0
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Jan 3, 2023 at 20:51 history edited David G. Stork CC BY-SA 4.0
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Jan 3, 2023 at 17:55 comment added user64494 @flinty: I have strong doubts about trig polynomials instead of polynomials .
Jan 3, 2023 at 17:52 comment added flinty Quantification over functions is second-order logic. Mathematica's theorem prover works only for a subset of first-order logic. However, it may be possible to answer this if you restrict your $f$ to a certain class of functions parametrized by a finite set of parameters, e.g a truncated Taylor series.
Jan 3, 2023 at 10:20 review Close votes
Jan 8, 2023 at 3:06
Jan 3, 2023 at 10:07 answer added Daniel Huber timeline score: 5
Jan 3, 2023 at 9:56 comment added user64494 This is math, not Mathematica. At the present and in the near future ForAll and Exists do not deal with functions as variables.
Jan 3, 2023 at 8:29 comment added David G. Stork I'm interested in using existential quantifiers (ForAll, Exists, ...) and such to "prove" this relation in Mathematica. In another context: Reduce[\!(* SubscriptBox["[ForAll]", RowBox[{"s", ",", RowBox[{"s", "[Element]", TemplateBox[{}, "Reals"]}]}]](* SubscriptBox["[Exists]", RowBox[{"t", ",", RowBox[{"t", "[Element]", TemplateBox[{}, "Reals"]}]}]]\ s\ t\ < \ 0))]
Jan 3, 2023 at 6:52 comment added bmf David, I think that these links might be of interest to you since they are the demonstrations of the mean value theorem and intermediate value theorem. Not sure if you knew about these or not. Also, a clarifying question: are you trying to build a routine from scratch that does the trick or do you want to use built-in functions?
Jan 3, 2023 at 6:12 history edited David G. Stork CC BY-SA 4.0
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Jan 3, 2023 at 6:06 history asked David G. Stork CC BY-SA 4.0