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Thread: Binomial series #2 stuck!

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

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

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


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

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

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


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

    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.

    $\displaystyle f(x) = \frac{5}{(1+\frac{x}{10})^4}$
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  4. #4
    o_O
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    Binomial series are given by: $\displaystyle (1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$$\displaystyle {\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 $\displaystyle k \in \mathbb{R}$ and if $\displaystyle |y| < 1$. Here, $\displaystyle y = \tfrac{x}{10}$ and $\displaystyle k = -4$ .

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

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

    _______________

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

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

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

    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: $\displaystyle (1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$$\displaystyle {\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 $\displaystyle k \in \mathbb{R}$ and if $\displaystyle |y| < 1$. Here, $\displaystyle y = \tfrac{x}{10}$ and $\displaystyle k = -4$ .

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

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

    _______________

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

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

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

    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: $\displaystyle \frac{4 \cdot 5 \cdot 6 \cdots (n+3)}{n!}$

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

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