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[CalcII/DiffEq] Closed form expression for f(x) which the series converges

  1. Nov 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Find a closed form expression for the function f(x) which the power series Σn=0..∞ n(-1)nxn+1 converges to and determine the values of x for which f(x) equals the given power series.

    2. Relevant equations

    3. The attempt at a solution
    I'm actually not sure how to start. First, what does it mean by "closed form expression"? Does that just mean an expression that isn't expressed as a series (with summation notation)?

    The only thing I can think of is writing the terms out. That's probably not right though. If anyone can give me a hint or nudge me in the right direction, I'd be extremely grateful. Thank you.
  2. jcsd
  3. Nov 21, 2014 #2
    A couple of examples of closed form expressions for power series are $$xe^x=\sum\limits_{k=0}^\infty\frac{x^{k+1}}{k!}$$ and $$\frac{x^2}{x+1}=\sum\limits_{k=0}^\infty(-1)^kx^{k+2}$$

    The left-hand sides there are the "closed forms". The idea is that you try to manipulate the sum you are given so that it looks like a power series that you "know". You may need to differentiate or take an antiderivative and/or possibly multiply/divide by a power of ##x## to make that happen.
  4. Nov 21, 2014 #3
    Oh, that makes a lot of sense. Thank you! I will attempt this now and post an update.
  5. Nov 21, 2014 #4
    Okay, I narrowed the possible forms to either be a variation of the arctan function or ln(1+x).

    It seems very similar to arctan, except the n isn't in the denominator. Is it even possible to invert the n? It doesn't seem like it...
  6. Nov 22, 2014 #5
    Okay, ignore the previous posts. I heeded gopher's advice and noticed that it looks similar to f(x) = 1/1-x.

    f(x) = 1/1-x = Σn=0..∞ xn
    f'(x) = 1/(1-x)2 = Σn=0..∞ nxn-1 = Σn=0..∞ nxnx-1
    -x2 * f'(x) = -x2/(1-x)2 = Σn=0..∞ -nxn+1 = Σn=0..∞ -nxnx

    Then, I believe my next step is to substitute -1x into the function:

    -x2/(1-(-x))2 = Σn=0..∞ -n(-x)n(-x) = Σn=0..∞ n(-1)nxn+1

    Thus, if I did this correctly, my expression should be: -x2/(1+x)2

    If this is correct, then I am assuming my next step would be to find the interval of convergence of the equivalent series to this expression. If anyone can verify this for me, I'd be extremely grateful.
    Last edited: Nov 22, 2014
  7. Nov 22, 2014 #6
    This looks mostly correct. The idea certainly is. I didn't check it completely, though, and the only part where I am concerned is when you got around to plugging in ##-x##. It seems like you may have lost a minus sign somewhere. Then again, I could be wrong.

    As for checking the interval of convergence, there is a theorem which I'm guessing (hoping?) you learned regarding the relationship between the interval of convergence of a power series and the intervals of convergence of the power series that you obtain by taking the term-by-term derivatives and antiderivatives. Essentially you don't gain or lose anything substantial when you do those things; you just need to check the endpoints. In the case of derivatives, you may lose endpoints, but you never gain them. For antiderivatives, it's the other way around.
  8. Nov 22, 2014 #7
    Thanks so much for having a look, gopher! After I plugged in -x, I got a (-x)n and a regular -x. I just multiplied the regular -x by the -1 that was originally there to get x (which I then combined with xn to get xn+1 after pulling out the (-1)n).

    Okay, so for the original power series, the interval of convergence was (-1,1). With the ratio test, the interval of the convergence appears to be the same after testing the endpoints x=1 and x=-1 with the test for divergence. So would it then be correct to conclude that the values of x for which f(x) equals the power series is between -1 and 1?
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