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Math Help - Taylor Series and shifting indices

  1. #1
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    Taylor Series and shifting indices

    For the truncated form of the Taylor series, the Nth Taylor polynomial is given in our notes as

    T_{N}(x) = \sum_{n = 0}^N \frac{f^{(n)}(0)}{n!} \cdot  x^{n}

    But then later, he defines T_{3}(x) for f(x)=\sin x

    as x-\frac{x^{3}}{6}

    Should it not be expanded to three terms, or is the notation signifying that it should be expanded until n, rather than N, reaches 3?

    Also, I have a question about shifting indices. Our notes give this equality

    \sum_{n = 1}^\infty \frac{(-1)^{n}4^{n}}{n!} = \sum_{n = 1}^\infty \frac{(-4)^{n}}{n!} = \sum_{n = 0}^\infty \frac{(-4)^{n}}{n!} - 1 = e^{-4} - 1 ,

    but is it not:

    \sum_{n = 1}^\infty \frac{(-4)^{n}}{n!} = \sum_{n = 0}^\infty \frac{(-4)^{n}}{n!} - \frac{4}{n+1}?

    Thanks
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  2. #2
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    For the first, notice that all even-numbered terms cancel since \sin(0)=0. For the second, you're just adding the term corresponding to n=0, I don't really know what you did to get to the other expression.
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  3. #3
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    Oh ok I see now, his notes just said to take 1 away from n in the index and increase it in the summand, not very clear to be honest! Thank you for that
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