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Math Help - Binomial series #2 stuck!

  1. #1
    Senior Member mollymcf2009's Avatar
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    Binomial series #2 stuck!

    f(x) = \frac{5}{(1+\frac{x}{10})^4}

    = 5\left[1 + \frac{x}{10})^-4\right]


    = 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]

    But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    f(x) = \frac{5}{(1+\frac{x}{10})^4}

    = 5\left[1 + \frac{x}{10})^-4\right]


    = 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]

    But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!
    Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?
    Sorry...

    Here is the whole question:

    Use the binomial series to expand the function as a power series.

    f(x) = \frac{5}{(1+\frac{x}{10})^4}
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  4. #4
    o_O
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    Binomial series are given by: (1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2 {\color{white}.} \ +  \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots

    for any k \in \mathbb{R} and if |y| < 1. Here, y = \tfrac{x}{10} and k = -4 .

    So: g(x) = \left( 1 + \frac{x}{10}\right)^{-4}

    g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots + \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots

    _______________

    Let's look at the general term: \frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}

    Just looking at the expanded series, we see that with even n, the coefficient is positive and with odd n, the coefficient is negative.

    So, if we factor out the negative signs: \frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots

    etc.
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  5. #5
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by o_O View Post
    Binomial series are given by: (1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2 {\color{white}.} \ +  \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots

    for any k \in \mathbb{R} and if |y| < 1. Here, y = \tfrac{x}{10} and k = -4 .

    So: g(x) = \left( 1 + \frac{x}{10}\right)^{-4}

    g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots + \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots

    _______________

    Let's look at the general term: \frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}

    Just looking at the expanded series, we see that with even n, the coefficient is positive and with odd n, the coefficient is negative.

    So, if we factor out the negative signs: \frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots

    etc.
    Ok, I understand everything except where the 3!/3! got there.

    Thanks!!!!
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  6. #6
    o_O
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    Focus on the expression: \frac{4 \cdot 5 \cdot 6 \cdots (n+3)}{n!}

    We can simplify it by multiplying by "1": \frac{{\color{red} 1 \cdot 2 \cdot 3} \cdot 5 \cdot 6 \cdots (n+3)}{n! \cdot {\color{red}3!}} = \frac{(n+3)!}{n! \cdot 3!}  = \frac{(n+3)(n+2)(n+1)n!}{n! \cdot 3!} = \frac{(n+3)(n+2)(n+1)}{ 6}

    So you can see why I multiplied both top and bottom by 3! so that I can simplify that long product into a simple factorial.
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