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Math Help - term-by-term differentiation

  1. #1
    Super Member Anonymous1's Avatar
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    term-by-term differentiation

    Find the sum using term-by-term differentiation:

    f(x) = \frac{x^3}{1\cdot 3} + \frac{x^4}{2\cdot 4} + \frac{x^5}{3\cdot 5} + \frac{x^6}{4\cdot 6}...

    I just don't know how to deal with the 'n' term. Taking the third derivative I get, for n \in [3,\infty), (n-1)x^{n-3} = \frac{n-1}{1-x} = ...?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Anonymous1 View Post
    Find the sum using term-by-term differentiation:

    f(x) = \frac{x^3}{1\cdot 3} + \frac{x^4}{2\cdot 4} + \frac{x^5}{3\cdot 5} + \frac{x^6}{4\cdot 6}...

    I just don't know how to deal with the 'n' term. Taking the third derivative I get, for n \in [3,\infty), (n-1)x^{n-3} = \frac{n-1}{1-x} = ...?
    Here's my thought:

    f'(x)=x\sum_{n=1}^{\infty}\frac{x^n}{n}=x\sum_{n=0  }^{\infty}\frac{x^{n+1}}{n+1}

    The derivative of the sum (not including the x outside it) is \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}

    So the sum itself is \int\frac{1}{1-x}\,dx=-\ln(1-x) and it follows that f'(x)=-x\ln(1-x).

    Integrate that to find f(x).

    Note that f(x) will only be defined on [-1,1).
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  3. #3
    Senior Member bkarpuz's Avatar
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    Red face

    Quote Originally Posted by Anonymous1 View Post
    Find the sum using term-by-term differentiation:

    f(x) = \frac{x^3}{1\cdot 3} + \frac{x^4}{2\cdot 4} + \frac{x^5}{3\cdot 5} + \frac{x^6}{4\cdot 6}...

    I just don't know how to deal with the 'n' term. Taking the third derivative I get, for n \in [3,\infty), (n-1)x^{n-3} = \frac{n-1}{1-x} = ...?
    It is not hard to see that
    f(x)=\sum_{k=1}^{\infty}\frac{x^{k+2}}{k(k+2)}.
    Here we have a_{k}:=\frac{1}{k(k+2)} for all k\in\mathbb{N}.
    For this power series, we see that the radius of convergence is r:=\lim_{k\to\infty}\frac{a_{k}}{a_{k+1}}=\lim_{k\  to\infty}\frac{k(k+2)}{(k+1)(k+3)}=1.
    Hence \sum_{k=1}^{\infty}\frac{x^{k+2}}{k(k+2)} converges for all |x|<r=1, you may check the end points.
    Thus, this power series converges uniformly in [-r_{0},r_{0}] for any r_{0}\in[0,r)=[0,1), hence term-by-term differentiable (and integrable) on the same interval.
    As we may pick r_{0} arbitrarily closer to r=1, we get
    f^{\prime}(x)=\bigg(\sum_{k=1}^{\infty}\frac{x^{k+  2}}{k(k+2)}\bigg)^{\prime}
    ........ =\sum_{k=1}^{\infty}\bigg(\frac{x^{k+2}}{k(k+2)}\b  igg)^{\prime}
    ........ =\sum_{k=1}^{\infty}\frac{(k+2)x^{k+1}}{k(k+2)}
    ........ =\sum_{k=1}^{\infty}\frac{x^{k+1}}{k}
    for all |x|<r=1.

    See Uniform Convergence -- from Wolfram MathWorld
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