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Can any one give me the derivation of this series

  1. Mar 22, 2013 #1
    Hi PF,

    I have 2 doubts :

    1. Can any one give me the derivation of series below :

    X = 1 + 2n + 3 n^2 + 4 n^3 + ..................

    2. I can do this series reverse if the result is given, using Mc'laurin series, but then came doubt regarding derivation of Taylor series. How do we derive Taylor series equation actually.
     
  2. jcsd
  3. Mar 22, 2013 #2

    Simon Bridge

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    What do you mean by "derivation" of a series?
    $$X=1+2n+3n^2+4n^3+... = \sum_{i=0}^\infty (i+1)n^i$$
    ... under what conditions will the sum converge?

    The MacLauran series is a special case of the Taylor series.
    They work the same way - if you can do one, you can do the other.
    http://en.wikipedia.org/wiki/Taylor_series
     
  4. Mar 22, 2013 #3

    HallsofIvy

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    If you mean "find the sum", that is typically extremely difficult. I am puzzled by your use of "Taylor's" series to "find the series" from the sum. The Taylors series is a series of functions, not a numerical series as you have here.
     
  5. Mar 22, 2013 #4
    If I'm not mistaken, Dexterdev, you are trying to find a formula for the finite sum of the above-mentioned series? Sure this could be difficult. But just out of curiousity, why would you want to find the formula for this sum?
     
  6. Mar 22, 2013 #5
    I want the sum because this is appearing somewhere in engineering (switching theory I believe), my friend asked for the "derivation" of series sum of infinite series.

    She has its end result from the textbook ie 1/((1-n)^2) , provided |n|<1

    I did the reverse using Taylor series ie from 1/((1-n)^2) , I got back the series. but how to arrive it other way?
     
  7. Mar 22, 2013 #6

    Curious3141

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    Hints:

    What sort of series is ##1 + n + n^2 + n^3 + ...## for ##|n| < 1##?

    If you represent the above series as a function of n, say ##S(n)##, what is its derivative ##S'(n)##?

    Can you proceed from here?

    Another well known way is to compute ##nX(n)## (##X(n)## represents your original series), then figure out what happens when you take ##X(n) - nX(n)## term by term. That's called the method of differences.
     
    Last edited: Mar 23, 2013
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