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Limit of a series

  1. Apr 4, 2006 #1
    so there is a power series
    S 4^n z^(3n)
    and upper limit being infinity and lower limit being 0. (S means sigma)
    then my book says, ak = 4^ (k/3) if k = 0, 3, 6...
    and ak equals 0 otherwise.
    i dont get how they came up for the value of ak

    i might be missing something silly ...
    any clues?
    thanks!
     
  2. jcsd
  3. Apr 5, 2006 #2

    dextercioby

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    Well, the way you asked it

    [tex] S=\sum_{n=0}^{\infty} \left(4z^{3}\right)^{n} [/tex]

    exists iff [itex] |4z^{3}|<1 [/itex].

    Daniel.
     
  4. Apr 5, 2006 #3

    Galileo

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    They are comparing it to the general form of a power series:
    [tex]\sum_{n=0}^\infty a_n z^n=a_0+a_1z+a_2z^2+...[/tex]

    You have
    [tex]\sum_{n=0}^\infty 4^nz^{3n}=1+4z^3+4^2z^6+...[/tex]

    Now what is the coefficient of [tex]z^k[/tex] in the given power series? (It's clearly zero if k is not a multiple of 3).
     
  5. Apr 5, 2006 #4
    since you seem to have a good understanding on power series and such ...
    could you also tell me how they expand rational functions?
    like i know 1/(1-x) = 1 +x +x^2/2! ....
    but how did they come up with that?

    and any other interesting rules about power series expansion of rational functions?
     
  6. Apr 5, 2006 #5
    you meant 1+x+x^2...

    by binomial theorem ?
     
  7. Apr 5, 2006 #6
    oops yeah ... there is no division by n factorial involved ....

    man i forgot all that power series and stuff ....

    thanks!
     
  8. Apr 6, 2006 #7

    Galileo

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