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For f(x), find a power series representation centered at the given value of a and determine the radius of convergence.

$$ f(x) = \frac {4x} {x^2-2x-3} ; a=0.$$

How would i begin this?
And what does it mean at the given value of a?
Thanks!

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  • $\begingroup$ Do you know what it means to expand a function into a power series around a point? (It gives you the point; $x=0$...) $\endgroup$
    – blue
    Mar 27, 2014 at 22:23
  • $\begingroup$ I just know about a function to power series, but not at a given point $\endgroup$
    – user136088
    Mar 27, 2014 at 22:25
  • $\begingroup$ "At a point $a$" just means the series should be of the form $\sum a_n(x-a)^n$. $\endgroup$
    – OR.
    Mar 27, 2014 at 22:25
  • $\begingroup$ Does that mean we sub the value in for a? $\endgroup$
    – user136088
    Mar 27, 2014 at 22:30

1 Answer 1

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Hint: Start with $$ \frac {4x} {x^2-2x-3} = \frac {4x}{(x-1)^2-4} =\frac {x-3+3(x+1)}{(x-3)(x+1)}. $$

details:

then you get to $$ \frac {4x} {x^2-2x-3} = \frac 1{x+1} + \frac 3{x-3} = \sum_{n=0}^\infty (-1)^nx^n + \sum_{n=0}^\infty -3^{-n}x^n. $$

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  • $\begingroup$ Clever - partial fractions to geometric series. $\endgroup$
    – gt6989b
    Mar 27, 2014 at 22:26
  • $\begingroup$ How about finding the radius of convergence? $\endgroup$
    – user136088
    Mar 27, 2014 at 22:41
  • $\begingroup$ Also in your last step, is it supposed to be 1/(x+1) not 1/(x-1) ? $\endgroup$
    – user136088
    Mar 27, 2014 at 22:52
  • $\begingroup$ @user136088: I made the appropriate correction $\endgroup$
    – mookid
    Mar 27, 2014 at 22:57
  • $\begingroup$ sorry would the radius of convergence be 1? $\endgroup$
    – user136088
    Mar 27, 2014 at 23:22

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