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Conversion of a function to a power series

  1. Mar 6, 2014 #1
    Given the function f(x) = 1/(3-x) it can be represented in a power series by 1/3∑(x/3)n but is there any restriction on saying ∑(x-2)n except for x = 2? In the first case, R = 3 but the second case, R = 1 and on different intervals (i.e. (-3,3) and the other is (1,3). I just simply used the expression for 1/(1-r) = ∑rn and let r = x - 2.

    So, any problems? I understand it's a less efficient method to solve this problem but I guess I'm just trying to see different ways to represent the same function.
  2. jcsd
  3. Mar 6, 2014 #2
    It's hard to understand what you're asking.
  4. Mar 6, 2014 #3
    Does f(x) = 1/(3-x) = (1/3)∑(x/3)n for (-3,3)?
  5. Mar 6, 2014 #4
    If the sum is taken for all nonnegative n, then yes. For the infinite series converges to a/(1 - r) where a is the first term and r is the common ratio with |r| < 1.
  6. Mar 7, 2014 #5


    Staff: Mentor

    MathewsMD, apparently I wasn't the only one who was confused by your question. How is this question related to what you asked about ##\sum (x - 2)^n##? For fixed x, that series is also a geometric series that converges if |x - 2| < 0; i.e., if 1 < x < 3. And x certainly can equal 2 here.
  7. Mar 7, 2014 #6
    Hi, sorry for the ambiguity. I was just in a rush yesterday and will try to clear things up now.

    So, I want to express the given function, ## f(x) = \frac {1}{x-2} ## as a power series.

    Now, ## \frac {a}{1-r} = ∑_{n=0}^∞ar^n ; -1 < r < 1##

    There is one simple method for solving this question, and that is simply dividing f(x) by 2 and then just rearranging. If you need more explanation for this part, just let me know. What I want to do is represent the function in a different way.

    Representing the function as ## f(x) = \frac {1}{1 - (x+3)} ; r = x + 3; a = 1, f(x) = ∑_{n=0}^∞ (x+3)^n ; -3 < x < 3##

    Now sorry if I my response is still a little convoluted or if my derivation seems trivial and useless. I would just like to know if this is incorrect for any reason. I haven't seen a power series changed in a similar form before, it's always just been dividing out the constant (e.g. 2 in this case) to put the denominator of f(x) in the form of ##1 - ax##. I know this is a much more useful method that will help with more advanced problems, but my main concern is just to understand the concept and figuring out different ways to represent these functions and to see if my understanding is correct. If I'm not mistaken, there are infinite ways to represent the same function, and I am just wondering if mine is correct. Thanks!
  8. Mar 7, 2014 #7


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    ##\frac {1}{1 - (x+3)}## isn't equal to ##\frac {1}{x-2}##. There are an infinite number of ways to represent the same function as a geometric series. You can expand around any x value. If they just say "power series" they probably want you to expand it around x=0, i.e. as a series in powers of x. To do that write ##\frac {1}{x-2}=\frac{-1/2}{1-(x/2)}##.
  9. Mar 7, 2014 #8
    Sorry, I meant ## f(x) = \frac {1}{2-x} ##
  10. Mar 7, 2014 #9


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    ##\frac {1}{1 - (x+3)}## isn't equal to ## \frac {1}{2-x} ## either. You've got to start getting the algebra right.
  11. Mar 7, 2014 #10
    Representing the function as ## f(x) = \frac {1}{1 - (x - 1)} ; r = x - 1; a = 1, f(x) = ∑_{n=0}^∞ (x-1)^n ; 0 < x < 2##

    Third time's the charm. My mistake :)

    Anything wrong with this? And yes, I understand what you did and it definitely makes sense solving with that method as well (it's more applicable to more advanced problems). I was just wondering if there's anything mathematically incorrect in my posted work?
  12. Mar 8, 2014 #11


    Staff: Mentor

    I'm really confused.

    You started of with the "given function" as f(x) = 1/(x - 2).

    In a later post, you said the "given function" was f(x) = 1/(3 - x).

    In your last post, you have f(x) = 1/(1 - (x - 1)), which is not equal to either of the functions above. 1/(1 - (x - 1)) = 1/(2 - x).

    So, what exactly is the function you're working with?
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