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Thread: An Underivable Function?

  1. #1 An Underivable Function? 
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    For the last few weeks, I've been trying to find the derivative of the function . However, so far, I've been making no progress. The only answer I managed to find was 0, and even then I had to permit -1! to exist in the equations.

    Lately, I've been thinking of trying to do it by allowing as and then trying to derive it by using the product rule.

    Would that work?


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  3. #2 Re: An Underivable Function? 
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    Quote Originally Posted by Liongold
    For the last few weeks, I've been trying to find the derivative of the function . However, so far, I've been making no progress. The only answer I managed to find was 0, and even then I had to permit -1! to exist in the equations.

    Lately, I've been thinking of trying to do it by allowing as and then trying to derive it by using the product rule.

    Would that work?
    The factorial operation (which is a series of subtractions and multipication) only works for integers, which results in an undefined derivative. There's something known as the "gamma function" that's popular on the subject, although I myself haven't looked into it, so unfortunately I can't help.


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  4. #3  
    Forum Radioactive Isotope MagiMaster's Avatar
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    The Gamma function is basically what you're looking for. It's defined as an integral, so it should be possible to take it's derivative directly. It's an integral from 0 to infinity though, so I don't remember off hand exactly how that'd work. (The best I can think of right now would be the difference of two limits.)

    Edit: Oh. The Wiki page gives the derivative further down the page. It's still defined as an integral, just a different one.
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  5. #4 Re: An Underivable Function? 
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    Quote Originally Posted by Liongold
    For the last few weeks, I've been trying to find the derivative of the function . However, so far, I've been making no progress. The only answer I managed to find was 0, and even then I had to permit -1! to exist in the equations.

    Lately, I've been thinking of trying to do it by allowing as and then trying to derive it by using the product rule.

    Would that work?
    No, it won't work.

    There is a reason that you cannot find the derivative of the function f(n)=n! directly. It is defined only for positive integers n, and hence the notion of the derivative does not make sense.

    Magimater is correct. What you need to so is look at the gamma function, which is an analytic function defined for all complex numbers except non-positive integers
    and which has the property that for positive integers n. (Edited to correct formula for gamma).

    http://en.wikipedia.org/wiki/Gamma_function

    Note however, that while the gamma function is differentiable and produces n! for non-negative integers n, that is is not unique in so doing. In fact if you add to the gamma function any analytic function that vanishes on the positive integers you get anothe analytic function that also gives n! on those points. So, for instance the meromorphic function is another function that also fits the bill.
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  6. #5  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Sorry, just a nit-pick, but . I have no idea why. That's just the way it was defined.
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  7. #6  
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    Quote Originally Posted by MagiMaster
    Sorry, just a nit-pick, but . I have no idea why. That's just the way it was defined.
    You are correct.

    Everything else that I said still stands.
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  8. #7  
    Forum Freshman jmd_dk's Avatar
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    The derivative of the function f(x) = x! is given by Psi(x+1)*x!, where Psi is the polygamma function. You can read about the polygamme function here: http://en.wikipedia.org/wiki/Polygamma_function
    Oh, and I used the program Mathcad to find the answer :wink:



    Just to show that this is actually true, I've done a test. First of, x! = G(x+1) (G should be the greek capital gamma). This allow os to use non-integers. The image speeks for it self:



    And yeh, I do know that 169.999999! is incorrect notation. But you know what I mean :wink: . The point is, the derivative seems to be very correct indeed
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  9. #8  
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    Quote Originally Posted by jmd_dk
    The derivative of the function f(x) = x! is given by Psi(x+1)*x!, where Psi is the polygamma function. You can read about the polygamme function here: http://en.wikipedia.org/wiki/Polygamma_function
    Oh, and I used the program Mathcad to find the answer :wink:



    Just to show that this is actually true, I've done a test. First of, x! = G(x+1) (G should be the greek capital gamma). This allow os to use non-integers. The image speeks for it self:



    And yeh, I do know that 169.999999! is incorrect notation. But you know what I mean :wink: . The point is, the derivative seems to be very correct indeed
    You have given the derivative of the gamma function, not of n!. x! is not defined unless x is a non-negative integer. The gamma function is defined more broadly, and . But there are other functions, with other derivatives that also satisfy that condition. The net result is that x! is not a meaningful expression without further qualification and since the function itself is not well-defined, neither is the derivative.

    The issue is not formulas so much as an understanding of the concept of a derivative.
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  10. #9  
    Forum Freshman jmd_dk's Avatar
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    Quote Originally Posted by DrRocket
    You have given the derivative of the gamma function, not of n!.
    Not true... I've given the derivative of the gamma function of (x+1), not just the gamma function (of x). You are pretty much right though... Before it gives much meaning to talk about a derivative, the function has to be continues. Therefore, we have to substitute the x! with Γ(x+1). This works, satisfied the question and do give us the derivative, if we imagine the x! as a continues function...
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