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Anyone recognize this series expansion?

  1. Apr 27, 2008 #1
    Anyone recognize this series expansion??

    [tex]1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{51t^4}{4!}+.....[/tex]

    I looks kind of like [itex]e^t[/itex] but i am not sure how to deal with it.

    Can I factor something... I kind of suck at these. Someone give me a hint.
     
  2. jcsd
  3. Apr 27, 2008 #2

    nicksauce

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    I don't quite see the pattern... how do you get from 27 to 51, and what comes next?
     
  4. Apr 27, 2008 #3
    Oh...yes that should be an 81

    [tex]
    1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{81t^4} {4!}+.....[/tex]
     
  5. Apr 27, 2008 #4

    nicksauce

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    So you have (3t)^0 / 0! + (3t)^1 / 1! + (3t)^2 /2! + (3t)^3 / 3! + ...
    Surely you can see what this function is?
     
  6. Apr 28, 2008 #5
    I can now! By the way what the Christ is 0! ??? I want to say that for some strange reason it is 1...... but I don't know why??
     
  7. Apr 28, 2008 #6

    cristo

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    0!=1, by definition.
     
  8. Apr 28, 2008 #7
    Yeah. That is what I thought.... though I thought that there was more to it than 'because the math gods said so.'

    But I'll take it if that's all there is to it:smile:
     
  9. Apr 28, 2008 #8

    Dick

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    The math gods said so for a good reason. You want the factorial function to satisfy n!=n*(n-1)!. If you put n=1, then you'd better define 0!=1. You can see you're in big trouble trying to define (-1)!. But that's ok. This is also related to the properties of the gamma function. gamma(n+1)=n!. And gamma(1)=1. gamma(0) is undefined, it's a pole of the gamma function. So we'd better leave (-n)! undefined.
     
  10. Apr 28, 2008 #9

    HallsofIvy

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    0!= 1 because I SAY SO!!
     
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