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How do I produce a function from a series?

  1. Nov 11, 2007 #1
    A simple example is 1+x+x^2=x^3+...=1/(1-x)

    I know that is equals 1/(1-x), but how do you arrive at that? Thanks.
     
  2. jcsd
  3. Nov 11, 2007 #2
    There are lots of ways. The analysis route would be to show that 1+x+...+x^n=(1-x^{n+1})/(1-x). If |x|<1, then x^{n+1} -> 0 so the limit becomes just 1/(1-x).
     
  4. Nov 11, 2007 #3

    rock.freak667

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    well 1+x+x^2=x^3+... is like a+ar+ar^2+...
    so that a=1 and r=x
    and well if they stated that |x|<1 so this would imply that |r|<1

    so then the sum to infinity(which is the sum of the infinite series1+x+x^2+x^3+...)

    would be [tex]S_\infty=\frac{a}{1-r}[/tex]

    which is [tex]\frac{1}{1-x}[/tex]
     
    Last edited: Nov 11, 2007
  5. Nov 12, 2007 #4
    sort of a generating function approach.

    define a function f by the series:

    f(x) = 1 + x + x^2 + x^3 + ...

    now take a look at x*f(x):

    xf(x) = x + x^2 + x^3 + ...

    add them together:

    f(x) - xf(x) = (1 + x + x^2 + x^3 + ... ) - (x + x^2 + x^3 + ... ) = 1.
    notice the terms cancel out. so f(x) - xf(x) = 1.

    factor out f(x) from the terms in the left-hand-side:

    f(x)(1-x) = 1.

    dividing both sides by (1-x) yields the result:

    f(x) = 1/(1-x).
     
    Last edited: Nov 12, 2007
  6. Nov 12, 2007 #5

    Gib Z

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    the simplest way is do use polynomial division =]
     
  7. Nov 12, 2007 #6
    This is doesn't tell you anything about the radius of convergence, which may or may not be relevant.
     
  8. Nov 12, 2007 #7
    true. i just think it's a nice example of being able to play with a series to find an explicit formula, though this isn't the the most telling of its nature.
     
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