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X^x, does this function have a name ?

  1. Apr 6, 2005 #1

    uart

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    Hi, the other day I was explaining to somebody the difference between a power function like x^a and an exponential function like a^x (it was in relation to differentiation). So they asked me "what about x^x" and I was stumped to give it a name.

    BTW, I was able to differentiate x^x ok, I got ( 1+ln(x) ) x^x which I'm pretty sure is correct, I just dont know if x^x has a commonly used name.
     
  2. jcsd
  3. Apr 6, 2005 #2
    Indeed your differentiation is correct. It doesn't have any sort of name that I know of (I've never seen it used for anything, either). A more useful representation is just

    [tex]e^{x\ln x},[/tex]

    of course.
     
  4. Apr 6, 2005 #3

    dextercioby

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    It doesn't have a name.The exponential form saves you from learning yet another concept:logarthmic differentiation.

    Daniel.
     
  5. Apr 6, 2005 #4

    uart

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    Doh of course. I don't know why I didn't think of that, particularly as I was explaining to this guy how to express a^x as [tex]e^{x\ln a}[/tex] (in order to differentiate it) at that very time. :blushing:
     
  6. Apr 6, 2005 #5

    cronxeh

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    unfortunately there is no indefinite integral for x^x
     
  7. Apr 6, 2005 #6
    Anectodically : the function [tex] x^x [/tex] is called "power tower of order 2"
     
  8. Apr 6, 2005 #7

    dextercioby

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    Power tower is defined for an infinity of exponents...So it's a bit improper.

    I'm sure that the function [itex] x^{x} [/itex] is Riemann integrable on a subdomain from R...

    Daniel.
     
  9. Apr 6, 2005 #8
    Yes, it is, of course (it's continuous on closed subintervals of [itex]\mathbb{R}^+[/itex], and as demonstrated above, differentiable too!). There's no closed-form antiderivative though.
     
  10. Apr 6, 2005 #9

    dextercioby

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    [tex] \int x^{x} \ dx [/tex] cannot be expressed in terms of "common" special functions...It it were,i'm sure Mathematica would have done it...

    Daniel.
     
  11. Apr 6, 2005 #10

    Zurtex

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    I loved it when I first worked out the minimum point of xx, really brought calc alive for me. I didn't have it set as an example I just realised as soon as I had learnt all the needed things that I could do it.
     
  12. Apr 6, 2005 #11
    Proof by Mathematica?
    j/k
     
  13. Apr 6, 2005 #12

    dextercioby

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    Yeah,go the integrator's webpage and type that function.The calculations are made with the latest version of the software.

    Daniel.
     
  14. Apr 7, 2005 #13

    Zurtex

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    Actually not true, I've now reported a series of integrations to Wolfram on things their online integrator should be able to integrate and can't. But things fairly obscure and I don’t think any people other than very bored ones like me will find it.
     
  15. Apr 7, 2005 #14

    dextercioby

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    Give an example.You can't contradict me with words,not in math,at least.

    Daniel.
     
  16. Apr 7, 2005 #15
    lambert function and product i know have something to do with x^x=y
     
  17. Apr 8, 2005 #16

    mathwonk

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    as to x^x, I usually called it george, but sometimes ozymandias.

    I have occasionally wondered what the name of the following number is as well:

    0.01001000100001000001.......

    i also have found simple partial fractions that mathematica would not do, but which i did myself with pencil and paper.

    More importantly to me, when I was teaching graduate algebra and galois theory i used mathematica to do some of the discriminant calculations for solutions formulas of cubics and quadratics, and found that sometimes mathematica just gave the wrong answer.

    It did not say it couldn't do it, or that I had enetered it wrong (usually the case) but simply gave a false answer.

    The errors I made in entering data were subtle, like omitting a space somewhere, but the upshot was that a naive student, who believed everything he got back, would have been misled by these incorrect responses, just as when using any calculator. So you always have to know how to check the validity of what you get from a calculator or computer.
     
    Last edited: Apr 8, 2005
  18. Apr 8, 2005 #17

    dextercioby

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    Reminds of my Maple.I had to reverse a fraction in an ODE so he could be able to solve it...:grumpy:

    A software is made by a human.Computers beat human mind for speed/efficiency.

    Daniel.
     
  19. Apr 8, 2005 #18

    Zurtex

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    It can compute the two following:

    [tex]\int \frac{1}{e^x + 2 e^{2 x} + 3 e^{3 x}} dx[/tex]

    [tex]\int \frac{x^2}{e^x + 2 e^{2 x} + 3 e^{3 x}} dx[/tex]

    But not:

    [tex]\int \frac{1 + x^2}{e^x + 2 e^{2 x} + 3 e^{3 x}} dx[/tex]
     
  20. Apr 8, 2005 #19

    dextercioby

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    :rofl: Yeah,it's a big black bug there...

    Mathematica latest version won't do this

    [tex] \int \frac{x+1}{e^{x}+2e^{2x}+3e^{3x}} \ dx [/tex]

    A 10 y.o. Maple returns proudly

    [tex] \int \frac{x+1}{e^x+2e^{2x}+3e^{3x}}dx=\allowbreak -\frac {1}{e^x}\ln \left( e^x\right) -\frac {2}{e^x}-\ln ^2\left( e^x\right)+\allowbreak i\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \arctan \frac {1}{2}\sqrt{2} [/tex]

    [tex]+\frac {1}{8}i\sqrt{2}\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \ln \left( 1+2e^x+3e^{2x}\right) -\allowbreak \frac {1}{4}\sqrt{2}\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \arctan \left( \frac {3}{2}e^x\sqrt{2}+\frac {1}2\sqrt{2}\right) [/tex]

    [tex]+\frac {1}{4}\sqrt{2}\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \arctan \frac {1}{2}\sqrt{2}+\allowbreak \frac {1}{2}\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \ln \left( 1+2e^x+3e^{2x}\right) [/tex]

    [tex]-\allowbreak i\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \arctan \frac {1}{2}\sqrt{2}-\frac {1}{8}i\sqrt{2}\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \ln \left( 1+2e^x+3e^{2x}\right) [/tex]

    [tex]-\allowbreak \frac {1}{4}\sqrt{2}\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \arctan \left( \frac {3}{2}e^x\sqrt{2}+\frac {1}{2}\sqrt{2}\right) +\frac {1}{4}\sqrt{2}\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \arctan \frac {1}{2}\sqrt{2}[/tex]

    [tex]+\allowbreak \frac {1}{2}\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \ln \left( 1+2e^x+3e^{2x}\right) -i\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \arctan \left( \frac {3}{2}e^x\sqrt{2}+\frac {1}{2}\sqrt{2}\right) [/tex]

    [tex]-\frac {1}{4}i\sqrt{2}\ \mbox{dilog}\left( 3\frac{e^x}{-1+i\sqrt{2}}\right) - \ \mbox{dilog}\left( 3\frac{e^x}{-1+i\sqrt{2}}\right) +\allowbreak \frac {1}{4}i\sqrt{2} \ \mbox{dilog}\left( -3\frac{e^x}{1+i\sqrt{2}}\right) - \ \mbox{dilog}\left( -3\frac{e^x}{1+i\sqrt{2}}\right)[/tex]

    [tex] -2\ln \left( e^x\right) +\allowbreak \ln \left( 1+2e^x+3e^{2x}\right) -\frac {1}{2}\sqrt{2}\arctan \frac {1}{4}\left( 2+6e^x\right) \sqrt{2}
    +i\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \arctan \left( \frac {3}{2}e^x\sqrt{2}+\frac {1}{2}\sqrt{2}\right)+ C' [/tex]


    :approve:

    Daniel.
     
    Last edited: Apr 8, 2005
  21. Apr 8, 2005 #20
    Quick, someone take the derivative and find out whether it's right. :tongue2:
     
    Last edited: Apr 8, 2005
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