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How to write this function in series form?

  1. Oct 2, 2011 #1
    How can I write this is series form?

    f(n) = [itex]\frac{z^n * s}{100^n}[/itex] + 2*[itex]\frac{z^(n-1) * s}{100^(n-1)}[/itex] + [itex]\frac{z^(n-2) * s}{100^(n-2)}[/itex] + [itex]\frac{z^(n-3) * s}{100^(n-3)}[/itex] + [itex]\frac{z^(n-4) * s}{100^(n-4)}[/itex]


    I stopped at this amount of terms arbitrarily, but really I want the series to stop adding terms when n is equal to x in "z^(n-x)" and "100^(n-x)". How can I say that in a general form?
     
  2. jcsd
  3. Oct 2, 2011 #2
    EDIT: it is supposed to be z^(n-1), 100^(n-1), z^(n-2) and 100^(n-2) etc..

    not 100(n-1), 100(n-2) etc.. as it shows up in the OP
     
  4. Oct 2, 2011 #3
    I don't even know if it is called series form, but I was just wondering if there was a more general way to write it? Like if n is equal to 100, I don't want to have to write out 100 terms of the equation so it would be simpler if there were a generalized form...
     
  5. Oct 2, 2011 #4

    Mute

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    Well, the general j^th term looks like

    [tex]s\frac{z^{n-j}}{100^{n-j}},[/tex]

    so you'll just need to sum over the range of j you want. However, in your original post you had a 2 in your second term - is that is really supposed to be there? If yes, should there be similar factors in front of the other terms? If so you will need to modify the above general term accordingly. If the term with the two is the only one like that, then summing over the above general term will leave you short a [itex]sz^{n-1}/100^{n-1}[/itex], so you'd have to add that separately.
     
  6. Oct 3, 2011 #5
    Thank you so much! And yes, the 2 was supposed to be there. So how can I find out how many terms it will take for my function in the OP to equal 100,000 if z=5 and s=500?
     
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