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Math Help - Taylor series expansion

  1. #1
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    Taylor series expansion

    Find the Taylor series for centered around .

    I've gotten pretty close but I don't know how to put it back into summation format. This is my working:











    So we have:

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

    So I know the summation form must have but I can't find the pattern for

    Help please, thanks!
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    something=\frac{(2n)!}{n!2^n}
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    something=\frac{(2n)!}{n!2^n}
    Hi how is it (2n)! ? Isn't that even?
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Hello!

    You need to find some kind of formula to:

    1\cdot 3 \cdot 5 \cdot 7 \cdot ... \cdot (2n-1)

    but, the above equals to: \frac{(2n)!}{n!2^n}

    Check this formula for n=4.
    Try to prove it! It is easy!

    By the way:


    {(2n)!}=1\cdot 2 \cdot 3 \cdot 4 \cdot ... \cdot (2n)
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  5. #5
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    yup that was exactly the formula for the multiplication of odd numbers however how do you prove it? in other words how did you get that 'formula'?

    Also:

    can't I have something like this:

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  6. #6
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    A product of even numbers, 2*4*6*8*...*(2n) can be written as (2*1)(2*2)(2*3)(2*4)*...(2*n). Now factor out those "2"s: (2*2*2*2...2)(1*2*3*4...*n)= 2^n n!.

    To simplify a product of odd numbers, 1*3*5*7*...*(2n-1), multiply and divide by the "missing" even numbers:
    \frac{1*2*3*4*5*6*7*8...*(2n-1)(2n)}{2*4*6*8*...(2n)}

    The numerator is now (2n)! and the denominator is 2^n n!.
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