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Math Help - Power Series

  1. #1
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    Power Series

    Find the power series expansion about the given point of the function:

    z^3 + 6z^2 - 4z -3 about z0 = 1

    Can someone show how to do this...

    Thanks in advance
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    f(z) = z^{3} + 6 z^{2} - 4 z - 3 = a_{3} (z-1)^{3} + a_{2} (z-1)^{2} + a_{1} (z-1) + a_{0} (1)

    Because the coefficient of z^{3} in (1) is 1 it will be a_{3} = 1 so that is...

    f(z) - (z-1)^{3} = 9 z^{2} - 7 z -2 (2)

    Because the coefficient of z^{2} in (2) is 9 it will be a_{2} = 9 so that is...

    f(z) - (z-1)^{3} - 9 (z-1)^{2}= 11z -11 (3)

    ... so that is a_{1} = 11 and a_{0}=0. The answer to Your question is then...

    f(z) = z^{3} + 6 z^{2} - 4 z - 3 = (z-1)^{3} + 9 (z-1)^{2} + 11 (z-1) (4)

    Kind regards

    \chi \sigma
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  3. #3
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    or rewrite the polynomial

     z^3 + 6z^2 - 4z -3 = (z-1+1)^3 + 6(z-1+1)^2 - 4(z-1+1) - 3

     [(z-1)^3 + 3(z-1)^2 + 3(z-1) + 1 ] + 6[(z-1)^2 + 2(z-1) + 1] - 4[(z-1) + 1] -3

     (z-1)^3 + (z-1)^2 (3+6) + (z-1) (3 + 12 - 4 ) + 1 + 6 -4 -3

     = (z-1)^3  + 9(z-1)^2 + 11(z-1)
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