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Math Help - Maclaurin series question

  1. #1
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    Maclaurin series question

    It seems like it's impossible to get the maclaurin series of sinx, cosx, and ln(x+1) to have a series index beginning at n=0. I'm confused because I think I was taught that taylor series supposedly start at n=0. ?
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  2. #2
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    Quote Originally Posted by Kaitosan View Post
    It seems like it's impossible to get the maclaurin series of sinx, cosx, and ln(x+1) to have a series index beginning at n=0. I'm confused because I think I was taught that taylor series supposedly start at n=0. ?
    I don't quite understand your confusion. By series index, do you mean the beginning value of the sum. Like for \sum_{n=0}^{\infty} you mean the n=0 part? Please explain your confusion a little more.

    Taylor series do always start with n=0 if my above assumption is true. Note that by the formula for generating every term in a Taylor series that when you apply the fact that 0!=1 and that f^{0}(a)=f(a) (the 0th derivative of f(x) is f(x)) then you are simply left with f(a), which can be 0 for a lot of things.
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    Will you please show me the maclaurin series of cosx, sinx, and ln(x+1) with "n" beginning at zero? I can't find it...
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    The maclaurin series of f is \sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!}\:x^n

    For f(x) = \cos x, f^{(2n+1)}(0) = 0 and f^{(2n)}(0) = (-1)^n

    Therefore the maclaurin series of cos x is \sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!}\:x^{2n}
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  5. #5
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    Quote Originally Posted by Kaitosan View Post
    Will you please show me the maclaurin series of cosx, sinx, and ln(x+1) with "n" beginning at zero? I can't find it...
    \sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}

    \cos x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}

    \ln x + 1 = \sum_{n=1}^\infty \frac{(-1)^{(n+1)}}{n}x^{n}

    and the only resaon that the last one starts at n = 1 is that when n= 0, the first term of the series is zero.
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  6. #6
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    Thanks guys, I was just confused about few things.
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