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Interval of Convergence/Power Series

  1. Dec 5, 2007 #1
    Hi everyone! I'm new here but I really hope you guys can be of some more help then my professor. We have our final exam next week (class ended today!) and the following 2 questions are on our review sheet. However, we NEVER covered this material. I'm terrible at "self taught" math so I'm having trouble here. My professor has office hours next week that I plan on attending but she did say specifically that we need to know everything on here. Okay so here they are:

    1.Find the interval of convergence for the series:
    Summation(1 to infinity) of
    (2^(k)) / ((2k)!) * x^(k)

    2.Write the first five nonzero terms in a power series for
    x / (2+3x^(3))
    Find the interval of convergence. Use these five terms to estimate:
    Integral from 0 to 0.1 (xdx) / (2+3x^(3))

    Okay so I'm figuring for the first one, you do the ratio test on the k terms so you end up with just x's but I'm not sure how you exactly calculate the interval. I see in my book something about setting up an inequality with x less than 1 and greater than -1 but is this standard?

    And with the second one, we've never done anything with power series so I'm completely at a loss. I see in my book again that you take the derivatives of f(x) to calculate the c's plugging in 0 for x but really again, I don't know if that's standard or just for the examples I've seen.

    So I'm pretty much at a huge loss - help!! :confused:

    Thank you :smile:
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 5, 2007 #2
    1. Hopefully you've covered the ratio test for series. The ratio test says that if

    [tex]\lim_{k\to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1[/tex]

    then the series converges (and otherwise the test is either inconclusive or the series diverges). You can extend this result to the notion of an interval of convergence. In your example, the a_k would be

    [tex]\frac{2^k} {(2k)! \cdot x^k}[/tex]

    If you take the ratio the (k+1)-th term and the k-th term (in absolute value) and set limit of this ratio to be less than 1, then you can solve for the values of x that make this inequality true (which will be an interval or a union of intervals). It then follows that (by the ratio test) your series will converge for any value of x in this interval. The only minor detail is to check the endpoints of the interval.

    Here's an example. Consider

    [tex]\sum_{k=1}^\infty (-1)^k \cdot \frac{(3x)^k}{k}[/tex]

    This series will converge if the ratio of the (k+1)-th term with the k-th term is less than 1. So for what values of x will this be true?

    [tex]\lim_{k\to\infty} \left|
    \frac{ (-1)^{k+1}\cdot \frac{(3x)^{k+1}}{k+1}}{ (-1)^k\cdot \frac{(3x)^k}{k} }
    \right| =
    \lim_{k\to\infty} \left| \frac{k}{k+1} \cdot (3x)\right| = |3x|

    Now set this limit to be < 1 and solve for x.

    [tex]|3x|< 1 \implies |x| < \frac{1}{3}[/tex]

    Thus for [tex]x\in (-1/3, 1/3)[/tex] we are guaranteed that the series converges. This is ALMOST the interval of convergence. We still need to check if the endpoints give us convergence. If you have x=(-1/3), then the original sequence is

    [tex]\sum_{k=1}^\infty \frac{(-1)^k [3 \cdot (-1/3)]^k}{k}
    = \sum_{k=1}^\infty \frac{1}{k}[/tex]

    which diverges (harmonic series), so x=-1/3 is not a point of convergence. On the other hand, if x=1/3, then

    [tex]\sum_{k=1}^\infty \frac{(-1)^k [3 \cdot (1/3)]^k}{k}
    = \sum_{k=1}^\infty \frac{(-1)^k}{k}[/tex]

    which converges (alternating harmonic series). Thus, your interval of convergence is (-1/3, 1/3].

    Remarks: if you get 0 for your limit, then it does not matter what the value of x, and so your interval of convergence is the entire set of real numbers. If your limit is larger than 1 somehow, then there is no interval of convergence.
    Last edited: Dec 5, 2007
  4. Dec 5, 2007 #3
    2. Power series essentially boil down to the following formula:

    [tex]\frac{1}{1-t} = \sum_{k=0}^\infty t^k[/tex]

    provided |t|<1 (it's essentially a geometric series whose ratio is t; and the sum of a geometric series is 1/(1-ratio) = 1/(1-t)).

    Thus, you can treat your expression as:

    [tex]\frac{x}{2+3x^3} = x \cdot \left( \frac{1}{2(1+\frac{3x^3}{2})}\right)
    = \frac{x}{2} \cdot \left( \frac{1}{1+\frac{3x^3}{2}}\right)
    = \frac{x}{2} \cdot \left( \frac{1}{1- (-\frac{3x^3}{2})}\right)[/tex]

    The idea is to manipulate your expression so that you have something which resembles the form [tex]\frac{1}{1-t}[/tex] (so that you can use the formula above), and everything else is a polynomial in x (in this just, we just have x/2). Take it from here. Just use the formula above with t = -(3x^3)/2, and then distribute the x/2 into the series. Once you have that, it should be easy to figure out the first 5 terms.
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