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Math Help - Taylor Series

  1. #1
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    Taylor Series

    I have come across the following in regards to a taylor series. I know what the solution is but not how it was arrived at.

    1-summation n=0->inf ((-1)^n*4^n*x^2n)/2n!

    It's the 1-summation part that has me stumped.
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  2. #2
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    Quote Originally Posted by p75213 View Post
    I have come across the following in regards to a taylor series. I know what the solution is but not how it was arrived at.

    1-summation n=0->inf ((-1)^n*4^n*x^2n)/2n!

    It's the 1-summation part that has me stumped.
    I have no idea what "1- summation" means! Is it possible that the "1" is just the problem number? Also "x^2n" could be (x^2)n or x^(2n)- I suspect the latter- and 2n! correctly is 2(n!) but I suspect you mean (2n)!

    If I am correct about what you mean then this is \sum_{n=0}^\infty \frac{(-1)^n 4^n x^{2n}}{(2n)!}= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(4x^2)^n= sin(4x^2)
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  3. #3
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    The equation you have shown is correct. It is the Taylor series for sine^2 x. By 1-summation I meant 1 minus summation.
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