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Math Help - Evaluate efficiently

  1. #1
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    Evaluate efficiently

    This is a question from my numerical analysis class. I apologize in advance if I'm in the wrong thread...

    Evaluate
    p(x) = 1 - x^3/3! + x^6/6! - x^9/9! + x^12/12! - x^15/15!
    as efficiently as possible. How many multiplications are necessary? Assume all coefficients have been computed and stored for later use.

    So I know the answer is 7 multiplications by looking in the back of the book. I'm almost positive Horner's Method is needed to show this since it's a big part of this section in my book...

    I thought to let y = x^3 and rewrite. I messed around with substitutions of this sort but could never get 7 multiplications. I'm actually not even sure if that's the way to go about it.

    If someone could show this, it will be greatly appreciated!! Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Evaluate efficiently

    Quote Originally Posted by jzellt View Post
    This is a question from my numerical analysis class. I apologize in advance if I'm in the wrong thread...

    Evaluate
    p(x) = 1 - x^3/3! + x^6/6! - x^9/9! + x^12/12! - x^15/15!
    as efficiently as possible. How many multiplications are necessary? Assume all coefficients have been computed and stored for later use.

    So I know the answer is 7 multiplications by looking in the back of the book. I'm almost positive Horner's Method is needed to show this since it's a big part of this section in my book...

    I thought to let y = x^3 and rewrite. I messed around with substitutions of this sort but could never get 7 multiplications. I'm actually not even sure if that's the way to go about it.

    If someone could show this, it will be greatly appreciated!! Thanks
    The Horner method for the computation of a a polynomial of the form...

    p(x)= c_{0}\ x^{n} + c_{1}\ x^{n-1}+...+c_{n} (1)

    ... consists in the computation of the n-th term of the recursive relation...

    p_{k+1}= x\ p_{k}+c_{k+1}\ ;\ p_{0}=c_{0} (2)

    In this case n=5 and in (2) You must use x^{3} instead of x...

    Kind regards

    \chi \sigma
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