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Taylor series integration

  1. Oct 15, 2008 #1

    Is there a quicker way to find the formula for the nth derivative of a function, instead of finding the first several derivatives and trying to find a pattern, and using that pattern to form the equation for the nth derivative?

    Also, is there a formula for the nth derivative of x^x? I cant find a formula for its derivatives so i cant find the polynomial. By the third derivative it starts to get complicated, and by the 5th i need to use 4 lines on a sheet of notebook paper just to write the derivative. Then i gave up because it got too cumbersome. when x^x is diffferentiated, it gives a x^x(1+lnx), and when that's differentiated, it gives a x^(x-1), then that gives a x^(x-1)*(1+lnx-1/x) Those terms keep producing more and more things, all the while the x^x(1+lnx)^n and x^(x-1)*(1+lnx-1/x)^n keep adding more and more things to the equation until it becomes a monster.

    maybe if i tried differentiating logx(y)=x, but i cant figure out how to do that in the first place. I have a feeling that it will be even more complicated, because implicit differentation would be required.

    Also im curious, is x^x a function one step higher than exponentation? Just as exponentation is a step higher than multiplication and multiplication is a step higher than addition.

    And, how do i integrate the series of a series? like if the x of a taylor polynomial is another taylor polynomial. If that integration is impossible, then how do i find the taylor polynomial that represents a taylor polynomial of a taylor polynomial?
  2. jcsd
  3. Oct 17, 2008 #2

    Gib Z

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    Homework Helper

    Not in any "useful" form. Of course, [tex] \frac{d^n}{dx^n}[/tex] does it quite well.

    Same answer.
    Multiplication is *repeated* addition, Exponentiation is repeated *multiplication.* Is this repeated exponentiation .. ? =]

    For a general function that would seem quite difficult and would most likely result in integrals with non -elementary derivatives. For some special function that may reduce to something "nice" though.
  4. Oct 21, 2008 #3
    There isn't anything special about x^x, when you see that x^x = e^(x ln x).
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