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Thread: Maclaurin series for square root (1+x)

  1. #1
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    Maclaurin series for square root (1+x)

    I attempted to find the maclaurin series for the function Square root of 1+x.

    F(0)=1 first term= 1
    F'(0)=1/2 second term= (1/2)x
    F''(0)=-1/4 Third term (-1/4)x^2
    F'''(0)=3/8 fourth term (3/8*3!) x^3
    F''''(0)=15/16 fifth (-15/16*4!) x^4
    F'''''(0)105/32 six (105/32*5!) x^5
    Therefore,
    f(x)= 1+ (1/2)x + (-1/4)x^2 + (3/8*3!) x^3 + (-15/16*4!) x^4 + (105/32*5!) x^5

    The problem is to find generalize term .
    I have ( ((-1)^(n-1)) * something *x^n) / ((2^n) * (n!))

    I cannot find that "something". because it exists as 1 for the first, second, and the third term , but then it increases to 3,15, 105
    so it increases by factor of 3,5,7....
    Help..
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  2. #2
    MHF Contributor chisigma's Avatar
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    The binomial series is in general defined as...

    $\displaystyle (1+x)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n}\cdot x^{n}$ (1)

    .... where...

    $\displaystyle \binom{\alpha}{n}= \frac{\alpha\cdot (\alpha-1)\cdot (\alpha-2)\dots (\alpha-n+1)}{n!}$ (2)

    For $\displaystyle \alpha = \frac{1}{2}$ we have...

    $\displaystyle (1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}\cdot x - \frac{1}{2\cdot 4}\cdot x^{2} + \frac{1\cdot 3}{2\cdot 4\cdot 6}\cdot x^{3} - \dots$ (3)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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