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It is often said, by way of intuitive explanation, that the derivative of a function at a point is the slope of the line that “best fits” the function through that point. Can this be pressed into a mathematically-rigorous definition, somewhat along the lines of the prisoners’ dilemma?

That is, two parties, who cannot communicate with each other, are given the same function and the same point, and told that if they both come up with the same line through that point that is best relative to the function, they will be set free, otherwise they will both have to serve time.

Would it then be some kind of theorem in Probability/Logic/Game Theory, that they would, if acting rationally and intelligently, come up with the tangent line “almost surely”?

Regards,

Mike Jones

2.Jun.2011 (Beijing time)

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    Let's name the players ε and δ.2011-06-02
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    Related? http://en.wikipedia.org/wiki/Axiom_of_determinacy2011-06-02
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    @Dan: No, not really.2011-06-02
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    Andres, I'm not so sure. AD and AC are incompatible and the latter yields immeasurable sets etc. Is AD enough for calculus? I don't know.2011-06-02
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    @Dan: I work on AD, I am not just saying.2011-06-02
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    Okay, I believe you. I spent the last ten minutes searching for a connection and I can't find one. I'm only vaguely familiar with AD and I'm still curious to know exactly why there is no pertinent relationship to calculus. Is the language-game of ε and δ really the best answer to this question?2011-06-02
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    @Mike: I don't see what you gain from your definition. You are trying to make the silk purse of a rigorous definition from the sow's ear of 'best fit', and it can't work in my opinion.2011-06-02
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    I think your definition depends very heavily on the players already having the proper notion of derivative. If someone would give me the function f(x) = 0 if |x|<1 and f(x) = x otherwise and the point (x,y)=(0,0), I might guess that anyone without mathematical education would think the line y=x is the best fit, as opposed to the tangent line y=0.2011-06-04
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    Thank you everybody for your comments. I'm going to accept the answer of gfes.2011-07-29

2 Answers 2

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As I stated in my comment I don't think your solution quite works. However, I think I just thought of a nice game-theoretical way of defining the derivative of a function:

Two players are both given a function $f$ and a point $(x,f(x))$. Both players then independently come up with a line through $(x,f(x))$, say $f_1$ and $f_2$.

Now the players announce which line they chose and both come up with some $\epsilon>0$ and the minimum of these is selected.

Then the final stage of our game begins: if $|f(x+\epsilon) - f_1(x+\epsilon)| = |f(x+\epsilon) - f_2(x+\epsilon)|$, the game ties and nothing happens. If $|f(x+\epsilon) - f_1(x+\epsilon)| < |f(x+\epsilon) - f_2(x+\epsilon)|$ player 1 wins and player 2 loses. If $|f(x+\epsilon) - f_1(x+\epsilon)| > |f(x+\epsilon) - f_2(x+\epsilon)|$ player 2 wins and player 1 loses.

To spice the game up a little one should of course introduce prison sentences or death penalties for the loser at this point, but I will leave that as an exercise to the reader.

Any proper game theorist would now wonder about the optimal strategy for both players, and unless I am gravely mistaken that would be choosing the line corresponding to the right-derivative of $f$ at $(x,f(x))$.

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    You've misstated your rules for determining the winner; you must mean something like comparing the errors between the line and the function for each player, not just comparing the values of the lines.2011-06-05
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    Oh, yes of course. I'll correct it immediately.2011-06-05
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You've only talked about the 1st derivative of a function, and the geometric way of looking at "the derivative" as a tangent line only works for the 1st derivative. It does come as possible to talk about the 2nd, 2.5th, pi-th, the -1th derivative (the integral), and other derivatives via [fractional calculus].1 I don't see how a game-theoretic definition would allow for anything else than the 1st derivative.

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    I don't know anything about fractional calculus, but for the n-th derivative (where n is just a natural number) couldn't one basically use the same approach as in the first derivative case, using n-th degree polynomials instead of lines?2011-06-05