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Math Help - power series

  1. #1
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    power series

    write a power series for f(x) = 3/(2-4x). write the series in closed form. give the interval of convergence.

    3/(2-4x)

    = (3/2)[1/(1-2x)]

    sum n=o to oo (3/2)(2^n)(x^n)

    how do i get the interval of convergence?
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    is up to his old tricks again! Jhevon's Avatar
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    \frac {3}{2} \sum_{n=0}^{ \infty} (2x)^n

    We need not consider the constant  \frac {3}{2}

    By the Root Test, we have:

    \lim_{n \to \infty} \left| (2x)^n \right|^{ \frac {1}{n}} = \lim_{n \to \infty} \left|2x \right|

    The series converges for \left| 2x \right| = 2 \left| x \right| < 1 \Rightarrow |x| < \frac {1}{2}

    So the radius of convergence is \frac {1}{2}

    Test the endpoints:

    If x = \frac {1}{2}

    \Rightarrow \sum (2x)^n = \sum 1^n which diverges by the test for divergence

    If x = - \frac {1}{2}

    \Rightarrow \sum (2x)^n = \sum (-1)^n which diverges by the test for divergence

    so the Interval of convergence is \left( - \frac {1}{2} , \frac {1}{2} \right)
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    how did the exponent 1/n get there?

    oh never mind its root test not ratio...sorry. we werent taught to use that test so i it didnt come to my mind at first...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    how did the exponent 1/n get there?
    i used the root test, you always put an exponent 1/n.

    you would get the same result using the ratio test, but since the exponent of 2x was just n, i decided to use the root test to get rid of the power, instead of going through the root test and having to form a fraction and all that stuff

    the ratio test will give the same result, now that you know the procedure to find the IOC, try it using the ratio test
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    Here is a weaker version of the root canal test. The strong version is based on the notion of a "limsup" which is not convered in any Calculus course.

    Given a series \sum_{n=1}^{\infty}a_n. And given the limit L = \lim |a_n|^{1/n} (if it exists or infinite). We have the following cases:

    If L<1 then the series converges absolutely.

    If L>1 \mbox{ or }L=+\infty then the series diverges.

    If L=1 the test doth not work.

    This is a beautiful test, which should be taken advantage of as much as possible. I like it more than the ratio test. (In fact it is stronger!)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is a weaker version of the root canal test. The strong version is based on the notion of a "limsup" which is not convered in any Calculus course.
    except in advanced calculus


    This is a beautiful test, which should be taken advantage of as much as possible. I like it more than the ratio test. (In fact it is stronger!)
    lol, why call it the "root canal" test then?

    i always have trouble applying it when factorials are involved, i think i did it once though.
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    Quote Originally Posted by Jhevon View Post
    except in advanced calculus

    lol, why call it the "root canal" test then?

    i always have trouble applying it when factorials are involved, i think i did it once though.
    I found the following useful. (Chapter 12)

    1) \lim (n!)^{1/n} = +\infty
    2) \lim \frac{1}{n}  (n!)^{1/n} = \frac{1}{e}

    I can derive it. But I remember I derived it for you before.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I found the following useful. (Chapter 12)

    1) \lim (n!)^{1/n} = +\infty
    2) \lim \frac{1}{n}  (n!)^{1/n} = \frac{1}{e}

    I can derive it. But I remember I derived it for you before.
    Yes, i remember. i believe we derived it by the Ratio test though
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