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I General Nth Derivative

  1. Mar 6, 2016 #1
    I'm very interesting in functions of the nature:

    [tex]f(x) = x^{x}[/tex]
    [tex]f(x) = x^{x^{x}}[/tex]
    and so on. I believe these are called tetrations? Regardless, I sought to generalize the nth derivative of [itex]f(x)=x^x[/itex] and it is proving to be difficult.

    First I tried just repeatedly differentiating until I could see a pattern:

    [tex] f'(x) = x^{x}\left(1+\ln(x)\right)[/tex]
    [tex]f''(x) = x^{x-1}\left(1+x+2x\ln(x) + x\ln^{x}(x)\right) [/tex]
    [tex]f'''(x) = x^{x-2}\left( -1+3x+x^{2} +3x(1+x)\ln(x)+3x^{2}\ln^{2}(x)+x^{2}\ln^{3}(x)\right) [/tex]
    [tex]f^{4}(x) = x^{x-3}\left( 2-x+6x^{2}+x^{3}+4x(-1+3x+x^{2})\ln(x)+6x^{2}(1+x)\ln^{2}(x)+4x^{3}\ln^{3}(x)+x^{3}\ln^{4}(x) \right) [/tex]

    I couldn't seem to find a pattern here. So I tried representing [itex] f(x)=x^{x} [/itex] as a power series:

    [tex] f(x) = x^{x} = e^{x\ln(x)} = \sum_{n=0}^{\infty} \frac{1}{n!}x^n\ln^{n}(x) [/tex]

    Then maybe taking the derivatives of the power series could lead me to an easier pattern. After I took the first derivative I felt that it wasn't the case as:

    [tex] f'(x) = \sum_{n=0}^{\infty} \frac{1}{(n-1)!}\left(1+\ln(x)\right)x^{n-1}\ln^{n-1}(x) [/tex]

    Which looks to be at a first glance more complex. Putting it through mathematica kind of left me with the same feeling of getting no where.

    Is there anything I should be considering? Any methods that I could employ to solve my problem? I don't know much about higher level maths (I'm currently doing Calc III) .
     
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  3. Mar 6, 2016 #2

    Simon Bridge

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  4. Mar 6, 2016 #3
    I'm interesting personally in a formula that would give me any derivative. Say I wanted the 10th derivative for instance. Of course you could manually take d/dx 10 times, but I think a formula would be a nice discovery.
     
  5. Mar 6, 2016 #4

    Simon Bridge

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    Oh I getcha...
    I'd try plotting the successive analytic derivatives to see if the curves show any patterns.
    The alternative would be to prove there wasn't one.

    You are getting ##f^{(n)} = x^{x-n+1}\left(\sum_{i=0}^n P_i^{n-1}(x) \ln^i|x|\right)## ... something... where ##P_i^j(k)## is the ith polynomial in k of degree j.
    ... except the pattern breaks for n=4 with P^3,P^6,P^3,P^6,P^3 ... but this may point to a way to find a pattern.

    You may want to get a computer to generate a lot of derivatives.
    Good luck.
     
  6. Mar 6, 2016 #5
    I've gotten to a point to where if I could generalize the nth derivative of

    [tex] f(x) = \ln^{n}(x) [/tex]

    I might be on a good path. This generalization looks messy though.
     
  7. Mar 6, 2016 #6

    Ssnow

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    Using the general product rule:

    ## D^{m} f(x)=D^{m}\left(\sum_{n=0}^{\infty}\frac{1}{n!}x^{n}\ln^{n}{x}\right)=\sum_{n=0}^{\infty}\frac{1}{n!}D^{m}\left(x^{n}\ln^{n}{x}\right)=##

    ##=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{i=0}^{m}\binom{m}{i}D^{i}(x^{n})D^{m-i}(\ln^{n}{x})##

    put it into a math program and it generates the ##m##-derivatives ...
     
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