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Find the Summation Notation and the Radius of Convergence

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the Summation Notation and Radius of Convergence of this series.

    5, x, 10, x, ....

    3. The attempt at a solution
    I don't know how did they come up with that equation.. But the summation seems right.. Can anyone tell me how did they arrive with that equation? I've tried using the common ratio but I don't know what to do after having the ratio..

    And for the radius of convergence, its L = lim (An+1)/(An) as n=> infinity..

    Using the summation notation, I ended up infinityinfinity over infinityinfinity. I thought its equal to 1.. ~_~

    Sorry, seems like the latex codes now isn't the same as the latex codes before. I don't know how to use the new latex codes..
     

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    Last edited: Jul 17, 2011
  2. jcsd
  3. Jul 17, 2011 #2

    HallsofIvy

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    If n is odd, [itex](-1)^n= -1[/itex] and if n is even, [itex](-1)^n= 1[/itex]. That means that if n is odd [itex](1+ (-1)^n)/2= 0[/itex] and if n is even, [itex](1+ (-1)^n)/2= 1[/itex]. Of course, if whether n is even or odd, n+ 1 is the opposite.

    So for any odd n, [itex]5^{(1+(-1)^n)/2}= 5^0= 1[/itex] and [itex]x^{(1+ (-1)^{n+1})/2}= x^1= x[/itex] while for n even, [itex]5^{(1+(-1)^n)/2}= 5^1= 5[/itex] and [itex]x^{(1+ (-1)^{n+1})/2}= x^0= 1[/itex]. That is, [itex]5^{(1+(-1)^n)/2}x^{(1+(-1)^n)/2}[/itex] just alternates 5 and x as your sequence does.

    As for finding the "radius of convergence", it should be obvious that this is NOT a power series and so does not have a "radius of convergence".

    (The LaTeX codes are the same as before- it just has picked up a bug so you need to click on your "refresh" button everytime you use LaTeX.)
     
  4. Jul 17, 2011 #3
    I know that, no need to elaborate. I'm just wondering how can I come up with that equation if you only have the values "5, x, 10, x, ...." as a given.
    Uhm, I don't get it. Why is this not a power series?
     
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